Mixing
Trigonometric equation $sin2x=cos x$
What is the number of solutions to
$$sin2x=cos x$$
on the interval $[0,3pi]$
What I tried here is:
$sin2x=cos x\2sin xcos x=cos x$
dividing this by $2cos x$ I get
$sin x={1over2}$
And from here I know
$x={piover6}+2kpi$
And looking in the interval I can only find 2 solutions, $xin{{piover6},{25piover6}}$
But looking at the results, there should be 7 results, what am I missing? And what should I do to get these results
algebra-precalculus trigonometry
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
add a comment |
What is the number of solutions to
$$sin2x=cos x$$
on the interval $[0,3pi]$
What I tried here is:
$sin2x=cos x\2sin xcos x=cos x$
dividing this by $2cos x$ I get
$sin x={1over2}$
And from here I know
$x={piover6}+2kpi$
And looking in the interval I can only find 2 solutions, $xin{{piover6},{25piover6}}$
But looking at the results, there should be 7 results, what am I missing? And what should I do to get these results
algebra-precalculus trigonometry
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
6
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
5
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
2
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
1
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49
add a comment |
What is the number of solutions to
$$sin2x=cos x$$
on the interval $[0,3pi]$
What I tried here is:
$sin2x=cos x\2sin xcos x=cos x$
dividing this by $2cos x$ I get
$sin x={1over2}$
And from here I know
$x={piover6}+2kpi$
And looking in the interval I can only find 2 solutions, $xin{{piover6},{25piover6}}$
But looking at the results, there should be 7 results, what am I missing? And what should I do to get these results
algebra-precalculus trigonometry
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
What is the number of solutions to
$$sin2x=cos x$$
on the interval $[0,3pi]$
What I tried here is:
$sin2x=cos x\2sin xcos x=cos x$
dividing this by $2cos x$ I get
$sin x={1over2}$
And from here I know
$x={piover6}+2kpi$
And looking in the interval I can only find 2 solutions, $xin{{piover6},{25piover6}}$
But looking at the results, there should be 7 results, what am I missing? And what should I do to get these results
algebra-precalculus trigonometry
algebra-precalculus trigonometry
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
asked Nov 21 '18 at 22:51
AleksaAleksa
33612
33612
6
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
5
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
2
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
1
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49
add a comment |
6
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
5
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
2
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
1
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49
6
6
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
5
5
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
2
2
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
1
1
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49
add a comment |
1 Answer
1
active
oldest
votes
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $sin x = frac{1}{2}$. Also,
$$frac{25pi}{6} > frac{18pi}{6} = 3pi$$
so $frac{25pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that
$$cos x = sinleft(frac{pi}{2} - xright)$$
by using the angle difference formula for sine
$$sin(alpha - beta) = sinalphacosbeta - cosalphasinbeta$$
Therefore,
$$sin(2x) = cos x$$
is equivalent to
$$sin(2x) = sinleft(frac{pi}{2} - xright)$$
When does $sintheta = sinvarphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $varphi = theta$ or $varphi = pi - theta$. It also occurs if $varphi$ is coterminal with $theta$ or $pi - theta$. Hence, $sintheta = sinvarphi$ if
$$varphi = theta + 2kpi, k in mathbb{Z}$$
or
$$varphi = pi - theta + 2mpi, m in mathbb{Z}$$
At the risk of obscuring the symmetry argument, you could write that $sintheta = sinvarphi$ if
$$varphi = (-1)^ntheta + npi, n in mathbb{Z}$$
With that in mind, let's solve the equation.
begin{align*}
sin(2x) & = cos x\
sin(2x) & = sinleft(frac{pi}{2} - xright)
end{align*}
Hence,
begin{align*}
2x & = frac{pi}{2} - x + 2kpi, k in mathbb{Z} & 2x & = pi - left(frac{pi}{2} - xright) + 2mpi, m in mathbb{Z}\
3x & = frac{pi}{2} + 2kpi, k in mathbb{Z} & 2x & = pi - frac{pi}{2} + x + 2mpi, m in mathbb{Z}\
x & = frac{pi}{6} + frac{2kpi}{3}, k in mathbb{Z} & x & = frac{pi}{2} + 2mpi, m in mathbb{Z}
end{align*}
We want solutions in the interval $[0, 3pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $sin(2x) = cos x$ has seven solutions in this interval.
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
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As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $sin x = frac{1}{2}$. Also,
$$frac{25pi}{6} > frac{18pi}{6} = 3pi$$
so $frac{25pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that
$$cos x = sinleft(frac{pi}{2} - xright)$$
by using the angle difference formula for sine
$$sin(alpha - beta) = sinalphacosbeta - cosalphasinbeta$$
Therefore,
$$sin(2x) = cos x$$
is equivalent to
$$sin(2x) = sinleft(frac{pi}{2} - xright)$$
When does $sintheta = sinvarphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $varphi = theta$ or $varphi = pi - theta$. It also occurs if $varphi$ is coterminal with $theta$ or $pi - theta$. Hence, $sintheta = sinvarphi$ if
$$varphi = theta + 2kpi, k in mathbb{Z}$$
or
$$varphi = pi - theta + 2mpi, m in mathbb{Z}$$
At the risk of obscuring the symmetry argument, you could write that $sintheta = sinvarphi$ if
$$varphi = (-1)^ntheta + npi, n in mathbb{Z}$$
With that in mind, let's solve the equation.
begin{align*}
sin(2x) & = cos x\
sin(2x) & = sinleft(frac{pi}{2} - xright)
end{align*}
Hence,
begin{align*}
2x & = frac{pi}{2} - x + 2kpi, k in mathbb{Z} & 2x & = pi - left(frac{pi}{2} - xright) + 2mpi, m in mathbb{Z}\
3x & = frac{pi}{2} + 2kpi, k in mathbb{Z} & 2x & = pi - frac{pi}{2} + x + 2mpi, m in mathbb{Z}\
x & = frac{pi}{6} + frac{2kpi}{3}, k in mathbb{Z} & x & = frac{pi}{2} + 2mpi, m in mathbb{Z}
end{align*}
We want solutions in the interval $[0, 3pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $sin(2x) = cos x$ has seven solutions in this interval.
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $sin x = frac{1}{2}$. Also,
$$frac{25pi}{6} > frac{18pi}{6} = 3pi$$
so $frac{25pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that
$$cos x = sinleft(frac{pi}{2} - xright)$$
by using the angle difference formula for sine
$$sin(alpha - beta) = sinalphacosbeta - cosalphasinbeta$$
Therefore,
$$sin(2x) = cos x$$
is equivalent to
$$sin(2x) = sinleft(frac{pi}{2} - xright)$$
When does $sintheta = sinvarphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $varphi = theta$ or $varphi = pi - theta$. It also occurs if $varphi$ is coterminal with $theta$ or $pi - theta$. Hence, $sintheta = sinvarphi$ if
$$varphi = theta + 2kpi, k in mathbb{Z}$$
or
$$varphi = pi - theta + 2mpi, m in mathbb{Z}$$
At the risk of obscuring the symmetry argument, you could write that $sintheta = sinvarphi$ if
$$varphi = (-1)^ntheta + npi, n in mathbb{Z}$$
With that in mind, let's solve the equation.
begin{align*}
sin(2x) & = cos x\
sin(2x) & = sinleft(frac{pi}{2} - xright)
end{align*}
Hence,
begin{align*}
2x & = frac{pi}{2} - x + 2kpi, k in mathbb{Z} & 2x & = pi - left(frac{pi}{2} - xright) + 2mpi, m in mathbb{Z}\
3x & = frac{pi}{2} + 2kpi, k in mathbb{Z} & 2x & = pi - frac{pi}{2} + x + 2mpi, m in mathbb{Z}\
x & = frac{pi}{6} + frac{2kpi}{3}, k in mathbb{Z} & x & = frac{pi}{2} + 2mpi, m in mathbb{Z}
end{align*}
We want solutions in the interval $[0, 3pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $sin(2x) = cos x$ has seven solutions in this interval.
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
add a comment |
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $sin x = frac{1}{2}$. Also,
$$frac{25pi}{6} > frac{18pi}{6} = 3pi$$
so $frac{25pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that
$$cos x = sinleft(frac{pi}{2} - xright)$$
by using the angle difference formula for sine
$$sin(alpha - beta) = sinalphacosbeta - cosalphasinbeta$$
Therefore,
$$sin(2x) = cos x$$
is equivalent to
$$sin(2x) = sinleft(frac{pi}{2} - xright)$$
When does $sintheta = sinvarphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $varphi = theta$ or $varphi = pi - theta$. It also occurs if $varphi$ is coterminal with $theta$ or $pi - theta$. Hence, $sintheta = sinvarphi$ if
$$varphi = theta + 2kpi, k in mathbb{Z}$$
or
$$varphi = pi - theta + 2mpi, m in mathbb{Z}$$
At the risk of obscuring the symmetry argument, you could write that $sintheta = sinvarphi$ if
$$varphi = (-1)^ntheta + npi, n in mathbb{Z}$$
With that in mind, let's solve the equation.
begin{align*}
sin(2x) & = cos x\
sin(2x) & = sinleft(frac{pi}{2} - xright)
end{align*}
Hence,
begin{align*}
2x & = frac{pi}{2} - x + 2kpi, k in mathbb{Z} & 2x & = pi - left(frac{pi}{2} - xright) + 2mpi, m in mathbb{Z}\
3x & = frac{pi}{2} + 2kpi, k in mathbb{Z} & 2x & = pi - frac{pi}{2} + x + 2mpi, m in mathbb{Z}\
x & = frac{pi}{6} + frac{2kpi}{3}, k in mathbb{Z} & x & = frac{pi}{2} + 2mpi, m in mathbb{Z}
end{align*}
We want solutions in the interval $[0, 3pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $sin(2x) = cos x$ has seven solutions in this interval.
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, you also overlooked some of the solutions of the equation $sin x = frac{1}{2}$. Also,
$$frac{25pi}{6} > frac{18pi}{6} = 3pi$$
so $frac{25pi}{6}$ is not a valid solution.
Here is a different approach that should make it less tempting to divide. You can prove that
$$cos x = sinleft(frac{pi}{2} - xright)$$
by using the angle difference formula for sine
$$sin(alpha - beta) = sinalphacosbeta - cosalphasinbeta$$
Therefore,
$$sin(2x) = cos x$$
is equivalent to
$$sin(2x) = sinleft(frac{pi}{2} - xright)$$
When does $sintheta = sinvarphi$?
Consider the figure below:
Two directed angles have the same sine if the points where their terminal sides intersect the unit circle have the same $y$-coordinate, which occurs if $varphi = theta$ or $varphi = pi - theta$. It also occurs if $varphi$ is coterminal with $theta$ or $pi - theta$. Hence, $sintheta = sinvarphi$ if
$$varphi = theta + 2kpi, k in mathbb{Z}$$
or
$$varphi = pi - theta + 2mpi, m in mathbb{Z}$$
At the risk of obscuring the symmetry argument, you could write that $sintheta = sinvarphi$ if
$$varphi = (-1)^ntheta + npi, n in mathbb{Z}$$
With that in mind, let's solve the equation.
begin{align*}
sin(2x) & = cos x\
sin(2x) & = sinleft(frac{pi}{2} - xright)
end{align*}
Hence,
begin{align*}
2x & = frac{pi}{2} - x + 2kpi, k in mathbb{Z} & 2x & = pi - left(frac{pi}{2} - xright) + 2mpi, m in mathbb{Z}\
3x & = frac{pi}{2} + 2kpi, k in mathbb{Z} & 2x & = pi - frac{pi}{2} + x + 2mpi, m in mathbb{Z}\
x & = frac{pi}{6} + frac{2kpi}{3}, k in mathbb{Z} & x & = frac{pi}{2} + 2mpi, m in mathbb{Z}
end{align*}
We want solutions in the interval $[0, 3pi]$. As you should verify, we obtain a solution in this interval if $k = 0, 1, 2, 3, 4$ or $m = 0, 1$. Since these seven solutions are distinct, the equation $sin(2x) = cos x$ has seven solutions in this interval.
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
answered Nov 22 '18 at 12:50
N. F. TaussigN. F. Taussig
43.6k93355
43.6k93355
add a comment |
add a comment |
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6
Hint: In what situation can you not divide by $2cos(x)$?
– Nicholas Stull
Nov 21 '18 at 22:52
5
Don't divide, factor! Write it as $cos(x)[2sin(x)-1] = 0$ and use that if $ab=0$ then $a=0$ or $b=0$.
– Winther
Nov 21 '18 at 22:54
2
Also, you appear to be missing a solution to $sin(x) = 1/2$.
– Nicholas Stull
Nov 21 '18 at 22:56
Thanks on the hints, got it now! I've got 7 results as said. Weird that I missed something that obvious!
– Aleksa
Nov 21 '18 at 22:57
1
@Aleksa: it is a common mistake to transform an equation and forget under what conditions the transformation preserves the solution.
– Yves Daoust
Nov 21 '18 at 23:49