If $tan 2alpha cdot tan alpha = 1$, then what is $alpha$? Different methods give different answers.












7












$begingroup$



If $tan 2alphacdottan alpha = 1$, then what is $alpha$?




I tried two methods but got two different answers.



Method 1:



$$begin{align}
tan 2alphacdottan alpha = 1
&implies
frac{2tan alpha}{1 - tan ^2 alpha};tan alpha = 1 tag{1a}\[6pt]
&implies 2tan ^2alpha = 1 - tan ^2alpha tag{1b}\[6pt]
&implies tan ^2 alpha = frac{1}{3} tag{1c}\[6pt]
&implies tan alpha = pmfrac{1}{sqrt 3} tag{1d}\[6pt]
&implies tan alpha = tanleft(pmfrac{pi}{6}right) tag{1e}\[6pt]
&implies alpha = npi pm frac{pi}{6} ;text{where}; n in mathbb{Z} tag{1f}
end{align}$$



Method 2:
$$begin{align}
tan 2alpha cdot tan alpha = 1
&implies tan 2alpha = frac{1}{tan alpha} tag{2a}\[6pt]
&implies tan 2alpha = cot alpha tag{2b}\[6pt]
&implies tan 2alpha = tanleft(frac{pi}{2} - alpharight) tag{2c}\[6pt]
&implies 2alpha = npi + frac{pi}{2} - alpha text{?} tag{2d}\[6pt]
&implies alpha = frac{1}{3}left(npi + frac{pi}{2}right);text{where}; n in mathbb{Z} tag{2e}
end{align}$$



Which one is correct? Is there any mistake in the above solutions?






trigonometry







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$endgroup$












  • $begingroup$
    Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
    $endgroup$
    – N. F. Taussig
    Jan 14 at 2:23
















7












$begingroup$



If $tan 2alphacdottan alpha = 1$, then what is $alpha$?




I tried two methods but got two different answers.



Method 1:



$$begin{align}
tan 2alphacdottan alpha = 1
&implies
frac{2tan alpha}{1 - tan ^2 alpha};tan alpha = 1 tag{1a}\[6pt]
&implies 2tan ^2alpha = 1 - tan ^2alpha tag{1b}\[6pt]
&implies tan ^2 alpha = frac{1}{3} tag{1c}\[6pt]
&implies tan alpha = pmfrac{1}{sqrt 3} tag{1d}\[6pt]
&implies tan alpha = tanleft(pmfrac{pi}{6}right) tag{1e}\[6pt]
&implies alpha = npi pm frac{pi}{6} ;text{where}; n in mathbb{Z} tag{1f}
end{align}$$



Method 2:
$$begin{align}
tan 2alpha cdot tan alpha = 1
&implies tan 2alpha = frac{1}{tan alpha} tag{2a}\[6pt]
&implies tan 2alpha = cot alpha tag{2b}\[6pt]
&implies tan 2alpha = tanleft(frac{pi}{2} - alpharight) tag{2c}\[6pt]
&implies 2alpha = npi + frac{pi}{2} - alpha text{?} tag{2d}\[6pt]
&implies alpha = frac{1}{3}left(npi + frac{pi}{2}right);text{where}; n in mathbb{Z} tag{2e}
end{align}$$



Which one is correct? Is there any mistake in the above solutions?






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
    $endgroup$
    – N. F. Taussig
    Jan 14 at 2:23














7












7








7





$begingroup$



If $tan 2alphacdottan alpha = 1$, then what is $alpha$?




I tried two methods but got two different answers.



Method 1:



$$begin{align}
tan 2alphacdottan alpha = 1
&implies
frac{2tan alpha}{1 - tan ^2 alpha};tan alpha = 1 tag{1a}\[6pt]
&implies 2tan ^2alpha = 1 - tan ^2alpha tag{1b}\[6pt]
&implies tan ^2 alpha = frac{1}{3} tag{1c}\[6pt]
&implies tan alpha = pmfrac{1}{sqrt 3} tag{1d}\[6pt]
&implies tan alpha = tanleft(pmfrac{pi}{6}right) tag{1e}\[6pt]
&implies alpha = npi pm frac{pi}{6} ;text{where}; n in mathbb{Z} tag{1f}
end{align}$$



Method 2:
$$begin{align}
tan 2alpha cdot tan alpha = 1
&implies tan 2alpha = frac{1}{tan alpha} tag{2a}\[6pt]
&implies tan 2alpha = cot alpha tag{2b}\[6pt]
&implies tan 2alpha = tanleft(frac{pi}{2} - alpharight) tag{2c}\[6pt]
&implies 2alpha = npi + frac{pi}{2} - alpha text{?} tag{2d}\[6pt]
&implies alpha = frac{1}{3}left(npi + frac{pi}{2}right);text{where}; n in mathbb{Z} tag{2e}
end{align}$$



Which one is correct? Is there any mistake in the above solutions?






trigonometry







share|cite|improve this question











$endgroup$





If $tan 2alphacdottan alpha = 1$, then what is $alpha$?




I tried two methods but got two different answers.



Method 1:



$$begin{align}
tan 2alphacdottan alpha = 1
&implies
frac{2tan alpha}{1 - tan ^2 alpha};tan alpha = 1 tag{1a}\[6pt]
&implies 2tan ^2alpha = 1 - tan ^2alpha tag{1b}\[6pt]
&implies tan ^2 alpha = frac{1}{3} tag{1c}\[6pt]
&implies tan alpha = pmfrac{1}{sqrt 3} tag{1d}\[6pt]
&implies tan alpha = tanleft(pmfrac{pi}{6}right) tag{1e}\[6pt]
&implies alpha = npi pm frac{pi}{6} ;text{where}; n in mathbb{Z} tag{1f}
end{align}$$



Method 2:
$$begin{align}
tan 2alpha cdot tan alpha = 1
&implies tan 2alpha = frac{1}{tan alpha} tag{2a}\[6pt]
&implies tan 2alpha = cot alpha tag{2b}\[6pt]
&implies tan 2alpha = tanleft(frac{pi}{2} - alpharight) tag{2c}\[6pt]
&implies 2alpha = npi + frac{pi}{2} - alpha text{?} tag{2d}\[6pt]
&implies alpha = frac{1}{3}left(npi + frac{pi}{2}right);text{where}; n in mathbb{Z} tag{2e}
end{align}$$



Which one is correct? Is there any mistake in the above solutions?






trigonometry




trigonometry






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edited Jan 14 at 2:22









Blue

48.4k870154




48.4k870154










asked Jan 14 at 1:57









TokyToky

496




496












  • $begingroup$
    Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
    $endgroup$
    – N. F. Taussig
    Jan 14 at 2:23


















  • $begingroup$
    Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
    $endgroup$
    – N. F. Taussig
    Jan 14 at 2:23
















$begingroup$
Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
$endgroup$
– N. F. Taussig
Jan 14 at 2:23




$begingroup$
Type$implies$ to obtain $implies$; type $iff$ to obtain $iff$.
$endgroup$
– N. F. Taussig
Jan 14 at 2:23










3 Answers
3






active

oldest

votes


















4












$begingroup$

Note carefully for future reference that in solving equations you need to use $Leftrightarrow$ not $Rightarrow$, or to do something equivalent.



Both your answers are correct as far as they go, but incomplete.



In your first method, for $alpha=npi+fracpi6$ we check that
$$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
so this is a correct solution, while for $alpha=npi-fracpi6$ we have
$$tan2alphatanalpha=(-sqrt3)(-frac1{sqrt3})=1 ,$$
so this is also a correct solution.



For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$. The first gives
$$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
so this is a correct solution. The second gives $alpha=kpi+fracpi2$, so $tanalpha$ is undefined and this must be ruled out. The third gives
$$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
so this is also a correct solution. Therefore your second method should give an answer
$$alpha=tfrac13(npi+tfracpi2) ,quadhbox{where $n=3k$ or $n=3k+2$}.$$
The actual numbers obtained in this solution are then the same as in your first method.



Always check your answers if you start with an equation and derive (potential) solutions.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Your first method is correct.



    In the second method, notice that the equation
    $$tan 2alpha = frac{1}{tanalpha}$$
    is not valid when $alpha = frac{pi}{2} + npi, n in mathbb{Z}$ or when $alpha = frac{pi}{4} + frac{npi}{2}, n in mathbb{Z}$ or when $alpha = npi, n in mathbb{Z}$. Therefore, in your solution
    $$alpha = frac{pi}{6} + frac{npi}{3}, n in mathbb{Z}$$
    $n neq 3k + 1$, $k in mathbb{Z}$ since that would imply
    $$alpha = frac{pi}{6} + kpi + frac{pi}{3} = frac{pi}{2} + kpi, k in mathbb{Z}$$
    which is not valid.



    If we replace $n$ by $3k$, we obtain
    $$alpha = frac{pi}{6} + kpi, k in mathbb{Z}$$
    If we replace $n$ by $3k - 1$, we obtain
    $$alpha = frac{pi}{6} + kpi - frac{pi}{3} = -frac{pi}{6} + kpi, k in mathbb{Z}$$
    which agrees with your first solution.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      @Toky,



      There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
      $$
      dots -frac{pi}{6}, frac{pi}{6}, frac{3pi}{6}, frac{5pi}{6},dots
      $$
      Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.



      Hope this helps.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
        $endgroup$
        – Will R
        Jan 14 at 2:18










      • $begingroup$
        You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
        $endgroup$
        – GReyes
        Jan 14 at 2:33











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Note carefully for future reference that in solving equations you need to use $Leftrightarrow$ not $Rightarrow$, or to do something equivalent.



      Both your answers are correct as far as they go, but incomplete.



      In your first method, for $alpha=npi+fracpi6$ we check that
      $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
      so this is a correct solution, while for $alpha=npi-fracpi6$ we have
      $$tan2alphatanalpha=(-sqrt3)(-frac1{sqrt3})=1 ,$$
      so this is also a correct solution.



      For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$. The first gives
      $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
      so this is a correct solution. The second gives $alpha=kpi+fracpi2$, so $tanalpha$ is undefined and this must be ruled out. The third gives
      $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
      so this is also a correct solution. Therefore your second method should give an answer
      $$alpha=tfrac13(npi+tfracpi2) ,quadhbox{where $n=3k$ or $n=3k+2$}.$$
      The actual numbers obtained in this solution are then the same as in your first method.



      Always check your answers if you start with an equation and derive (potential) solutions.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Note carefully for future reference that in solving equations you need to use $Leftrightarrow$ not $Rightarrow$, or to do something equivalent.



        Both your answers are correct as far as they go, but incomplete.



        In your first method, for $alpha=npi+fracpi6$ we check that
        $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
        so this is a correct solution, while for $alpha=npi-fracpi6$ we have
        $$tan2alphatanalpha=(-sqrt3)(-frac1{sqrt3})=1 ,$$
        so this is also a correct solution.



        For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$. The first gives
        $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
        so this is a correct solution. The second gives $alpha=kpi+fracpi2$, so $tanalpha$ is undefined and this must be ruled out. The third gives
        $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
        so this is also a correct solution. Therefore your second method should give an answer
        $$alpha=tfrac13(npi+tfracpi2) ,quadhbox{where $n=3k$ or $n=3k+2$}.$$
        The actual numbers obtained in this solution are then the same as in your first method.



        Always check your answers if you start with an equation and derive (potential) solutions.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Note carefully for future reference that in solving equations you need to use $Leftrightarrow$ not $Rightarrow$, or to do something equivalent.



          Both your answers are correct as far as they go, but incomplete.



          In your first method, for $alpha=npi+fracpi6$ we check that
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is a correct solution, while for $alpha=npi-fracpi6$ we have
          $$tan2alphatanalpha=(-sqrt3)(-frac1{sqrt3})=1 ,$$
          so this is also a correct solution.



          For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$. The first gives
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is a correct solution. The second gives $alpha=kpi+fracpi2$, so $tanalpha$ is undefined and this must be ruled out. The third gives
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is also a correct solution. Therefore your second method should give an answer
          $$alpha=tfrac13(npi+tfracpi2) ,quadhbox{where $n=3k$ or $n=3k+2$}.$$
          The actual numbers obtained in this solution are then the same as in your first method.



          Always check your answers if you start with an equation and derive (potential) solutions.






          share|cite|improve this answer









          $endgroup$



          Note carefully for future reference that in solving equations you need to use $Leftrightarrow$ not $Rightarrow$, or to do something equivalent.



          Both your answers are correct as far as they go, but incomplete.



          In your first method, for $alpha=npi+fracpi6$ we check that
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is a correct solution, while for $alpha=npi-fracpi6$ we have
          $$tan2alphatanalpha=(-sqrt3)(-frac1{sqrt3})=1 ,$$
          so this is also a correct solution.



          For your second solution there are three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$. The first gives
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is a correct solution. The second gives $alpha=kpi+fracpi2$, so $tanalpha$ is undefined and this must be ruled out. The third gives
          $$tan2alphatanalpha=sqrt3frac1{sqrt3}=1 ,$$
          so this is also a correct solution. Therefore your second method should give an answer
          $$alpha=tfrac13(npi+tfracpi2) ,quadhbox{where $n=3k$ or $n=3k+2$}.$$
          The actual numbers obtained in this solution are then the same as in your first method.



          Always check your answers if you start with an equation and derive (potential) solutions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 2:20









          DavidDavid

          68.9k667130




          68.9k667130























              3












              $begingroup$

              Your first method is correct.



              In the second method, notice that the equation
              $$tan 2alpha = frac{1}{tanalpha}$$
              is not valid when $alpha = frac{pi}{2} + npi, n in mathbb{Z}$ or when $alpha = frac{pi}{4} + frac{npi}{2}, n in mathbb{Z}$ or when $alpha = npi, n in mathbb{Z}$. Therefore, in your solution
              $$alpha = frac{pi}{6} + frac{npi}{3}, n in mathbb{Z}$$
              $n neq 3k + 1$, $k in mathbb{Z}$ since that would imply
              $$alpha = frac{pi}{6} + kpi + frac{pi}{3} = frac{pi}{2} + kpi, k in mathbb{Z}$$
              which is not valid.



              If we replace $n$ by $3k$, we obtain
              $$alpha = frac{pi}{6} + kpi, k in mathbb{Z}$$
              If we replace $n$ by $3k - 1$, we obtain
              $$alpha = frac{pi}{6} + kpi - frac{pi}{3} = -frac{pi}{6} + kpi, k in mathbb{Z}$$
              which agrees with your first solution.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                Your first method is correct.



                In the second method, notice that the equation
                $$tan 2alpha = frac{1}{tanalpha}$$
                is not valid when $alpha = frac{pi}{2} + npi, n in mathbb{Z}$ or when $alpha = frac{pi}{4} + frac{npi}{2}, n in mathbb{Z}$ or when $alpha = npi, n in mathbb{Z}$. Therefore, in your solution
                $$alpha = frac{pi}{6} + frac{npi}{3}, n in mathbb{Z}$$
                $n neq 3k + 1$, $k in mathbb{Z}$ since that would imply
                $$alpha = frac{pi}{6} + kpi + frac{pi}{3} = frac{pi}{2} + kpi, k in mathbb{Z}$$
                which is not valid.



                If we replace $n$ by $3k$, we obtain
                $$alpha = frac{pi}{6} + kpi, k in mathbb{Z}$$
                If we replace $n$ by $3k - 1$, we obtain
                $$alpha = frac{pi}{6} + kpi - frac{pi}{3} = -frac{pi}{6} + kpi, k in mathbb{Z}$$
                which agrees with your first solution.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Your first method is correct.



                  In the second method, notice that the equation
                  $$tan 2alpha = frac{1}{tanalpha}$$
                  is not valid when $alpha = frac{pi}{2} + npi, n in mathbb{Z}$ or when $alpha = frac{pi}{4} + frac{npi}{2}, n in mathbb{Z}$ or when $alpha = npi, n in mathbb{Z}$. Therefore, in your solution
                  $$alpha = frac{pi}{6} + frac{npi}{3}, n in mathbb{Z}$$
                  $n neq 3k + 1$, $k in mathbb{Z}$ since that would imply
                  $$alpha = frac{pi}{6} + kpi + frac{pi}{3} = frac{pi}{2} + kpi, k in mathbb{Z}$$
                  which is not valid.



                  If we replace $n$ by $3k$, we obtain
                  $$alpha = frac{pi}{6} + kpi, k in mathbb{Z}$$
                  If we replace $n$ by $3k - 1$, we obtain
                  $$alpha = frac{pi}{6} + kpi - frac{pi}{3} = -frac{pi}{6} + kpi, k in mathbb{Z}$$
                  which agrees with your first solution.






                  share|cite|improve this answer











                  $endgroup$



                  Your first method is correct.



                  In the second method, notice that the equation
                  $$tan 2alpha = frac{1}{tanalpha}$$
                  is not valid when $alpha = frac{pi}{2} + npi, n in mathbb{Z}$ or when $alpha = frac{pi}{4} + frac{npi}{2}, n in mathbb{Z}$ or when $alpha = npi, n in mathbb{Z}$. Therefore, in your solution
                  $$alpha = frac{pi}{6} + frac{npi}{3}, n in mathbb{Z}$$
                  $n neq 3k + 1$, $k in mathbb{Z}$ since that would imply
                  $$alpha = frac{pi}{6} + kpi + frac{pi}{3} = frac{pi}{2} + kpi, k in mathbb{Z}$$
                  which is not valid.



                  If we replace $n$ by $3k$, we obtain
                  $$alpha = frac{pi}{6} + kpi, k in mathbb{Z}$$
                  If we replace $n$ by $3k - 1$, we obtain
                  $$alpha = frac{pi}{6} + kpi - frac{pi}{3} = -frac{pi}{6} + kpi, k in mathbb{Z}$$
                  which agrees with your first solution.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 14 at 11:08

























                  answered Jan 14 at 2:14









                  N. F. TaussigN. F. Taussig

                  44.2k93356




                  44.2k93356























                      1












                      $begingroup$

                      @Toky,



                      There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
                      $$
                      dots -frac{pi}{6}, frac{pi}{6}, frac{3pi}{6}, frac{5pi}{6},dots
                      $$
                      Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.



                      Hope this helps.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                        $endgroup$
                        – Will R
                        Jan 14 at 2:18










                      • $begingroup$
                        You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                        $endgroup$
                        – GReyes
                        Jan 14 at 2:33
















                      1












                      $begingroup$

                      @Toky,



                      There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
                      $$
                      dots -frac{pi}{6}, frac{pi}{6}, frac{3pi}{6}, frac{5pi}{6},dots
                      $$
                      Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.



                      Hope this helps.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                        $endgroup$
                        – Will R
                        Jan 14 at 2:18










                      • $begingroup$
                        You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                        $endgroup$
                        – GReyes
                        Jan 14 at 2:33














                      1












                      1








                      1





                      $begingroup$

                      @Toky,



                      There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
                      $$
                      dots -frac{pi}{6}, frac{pi}{6}, frac{3pi}{6}, frac{5pi}{6},dots
                      $$
                      Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.



                      Hope this helps.






                      share|cite|improve this answer









                      $endgroup$



                      @Toky,



                      There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
                      $$
                      dots -frac{pi}{6}, frac{pi}{6}, frac{3pi}{6}, frac{5pi}{6},dots
                      $$
                      Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.



                      Hope this helps.







                      share|cite|improve this answer












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                      answered Jan 14 at 2:12









                      GReyesGReyes

                      1,25915




                      1,25915












                      • $begingroup$
                        What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                        $endgroup$
                        – Will R
                        Jan 14 at 2:18










                      • $begingroup$
                        You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                        $endgroup$
                        – GReyes
                        Jan 14 at 2:33


















                      • $begingroup$
                        What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                        $endgroup$
                        – Will R
                        Jan 14 at 2:18










                      • $begingroup$
                        You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                        $endgroup$
                        – GReyes
                        Jan 14 at 2:33
















                      $begingroup$
                      What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                      $endgroup$
                      – Will R
                      Jan 14 at 2:18




                      $begingroup$
                      What value of $n$ gives $npipmdfrac{pi}{6}=dfrac{3pi}{6}=dfrac{pi}{2}$?
                      $endgroup$
                      – Will R
                      Jan 14 at 2:18












                      $begingroup$
                      You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                      $endgroup$
                      – GReyes
                      Jan 14 at 2:33




                      $begingroup$
                      You are right. Not all values of $n$ work for the first method. Thanks for pointing this out.
                      $endgroup$
                      – GReyes
                      Jan 14 at 2:33


















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