Mixing
Solving $2sinthetacostheta + sintheta = 0$
$begingroup$
The question is to solve the following question in the range $-pi le theta le pi$
$$2sinthetacostheta + sintheta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $pm2/3pi$ and the values when $sintheta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$begin{align}
2sinthetacostheta + sintheta &= 0 qquadtext{(square)} tag{1} \
4sin^2thetacos^2theta + sin^2theta &= 0 tag{2}\
4sin^2theta(1-sin^2theta) + sin^2theta &= 0 tag{3} \
4sin^2theta - 4sin^4theta + sin^2theta &= 0 tag{4} \
5sin^2theta - 4sin^4theta &= 0 tag{5}
end{align}$$
and then solving by substitution/the quadratic equation I get $sintheta = pmsqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
trigonometry
edited Feb 2 at 5:53
Blue
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
$endgroup$
add a comment |
$begingroup$
The question is to solve the following question in the range $-pi le theta le pi$
$$2sinthetacostheta + sintheta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $pm2/3pi$ and the values when $sintheta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$begin{align}
2sinthetacostheta + sintheta &= 0 qquadtext{(square)} tag{1} \
4sin^2thetacos^2theta + sin^2theta &= 0 tag{2}\
4sin^2theta(1-sin^2theta) + sin^2theta &= 0 tag{3} \
4sin^2theta - 4sin^4theta + sin^2theta &= 0 tag{4} \
5sin^2theta - 4sin^4theta &= 0 tag{5}
end{align}$$
and then solving by substitution/the quadratic equation I get $sintheta = pmsqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
trigonometry
edited Feb 2 at 5:53
Blue
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
$endgroup$
1
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06
add a comment |
$begingroup$
The question is to solve the following question in the range $-pi le theta le pi$
$$2sinthetacostheta + sintheta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $pm2/3pi$ and the values when $sintheta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$begin{align}
2sinthetacostheta + sintheta &= 0 qquadtext{(square)} tag{1} \
4sin^2thetacos^2theta + sin^2theta &= 0 tag{2}\
4sin^2theta(1-sin^2theta) + sin^2theta &= 0 tag{3} \
4sin^2theta - 4sin^4theta + sin^2theta &= 0 tag{4} \
5sin^2theta - 4sin^4theta &= 0 tag{5}
end{align}$$
and then solving by substitution/the quadratic equation I get $sintheta = pmsqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
trigonometry
edited Feb 2 at 5:53
Blue
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
$endgroup$
The question is to solve the following question in the range $-pi le theta le pi$
$$2sinthetacostheta + sintheta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $pm2/3pi$ and the values when $sintheta = 0$. Although I missed the early factorisation I don't know what I'm doing to actually arrive at an incorrect answer:
$$begin{align}
2sinthetacostheta + sintheta &= 0 qquadtext{(square)} tag{1} \
4sin^2thetacos^2theta + sin^2theta &= 0 tag{2}\
4sin^2theta(1-sin^2theta) + sin^2theta &= 0 tag{3} \
4sin^2theta - 4sin^4theta + sin^2theta &= 0 tag{4} \
5sin^2theta - 4sin^4theta &= 0 tag{5}
end{align}$$
and then solving by substitution/the quadratic equation I get $sintheta = pmsqrt(5)/2$ and $0$ but as this out of bounds for sin so cannot be the answer.
I know using Symbolab that this solution is correct for the quadratic I've generated, so I must be going wrong somewhere above after missing that factoristion. I feel like it is in the squaring step but not sure what would be wrong here...
Thanks a lot for your help.
trigonometry
trigonometry
edited Feb 2 at 5:53
Blue
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
edited Feb 2 at 5:53
Blue
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
edited Feb 2 at 5:53
Blue
49.6k870158
edited Feb 2 at 5:53
Blue
49.6k870158
edited Feb 2 at 5:53
Blue
49.6k870158
49.6k870158
asked Feb 1 at 15:03
JosephJoseph
304
asked Feb 1 at 15:03
JosephJoseph
304
asked Feb 1 at 15:03
JosephJoseph
304
304
1
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06
add a comment |
1
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06
1
1
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4sin^2 theta cos theta$. If you took the $sin theta$ to the other side before squaring to avoid the cross term, when you brought it back the $sin^2 theta$ should have a minus sign.
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
add a comment |
$begingroup$
I'll start by graphing this function x-axis is $theta / pi $ which shows the function is zero at 5 points.
begin{align}
2 cdot sin(theta)cos(theta) + sin(theta) & = 0 \
sin(theta) cdot (2cdotcos(theta)+1) & = 0
end{align}
So either $sin(theta) = 0$ or $2cdotcos(theta)+1 = 0 Rightarrow cos(theta) =
-0.5$
Considering each of these cases then $theta$ is $-pi$, $-dfrac{2}{3}pi$, $0$, $dfrac{2}{3}pi$ or $pi$
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
$endgroup$
add a comment |
$begingroup$
For $-pileqthetaleqpi$:
begin{align*}
2sin{theta}cos{theta}+sin{theta}&=0 \
sin{theta}big(2cos{theta}+1big)&=0 \
end{align*}
Thus, $sin{theta}=0$ or $2cos{theta}+1=0$ .
- If $sin{theta}=0$, then $thetain{-pi,0,pi}$ .
- If $2cos{theta}+1=0$, then $cos{theta}=-1/2<0$, then $thetain(-pi,-pi/2)cup(pi/2,pi)$ and thus $thetain{-2pi/3,2pi/3}$
The solution is $thetainbigg{-pi,-dfrac{2pi}{3},0,dfrac{2pi}{3},pibigg}$ .
answered Feb 1 at 15:14
El boritoEl borito
664216
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4sin^2 theta cos theta$. If you took the $sin theta$ to the other side before squaring to avoid the cross term, when you brought it back the $sin^2 theta$ should have a minus sign.
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
add a comment |
$begingroup$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4sin^2 theta cos theta$. If you took the $sin theta$ to the other side before squaring to avoid the cross term, when you brought it back the $sin^2 theta$ should have a minus sign.
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
add a comment |
$begingroup$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4sin^2 theta cos theta$. If you took the $sin theta$ to the other side before squaring to avoid the cross term, when you brought it back the $sin^2 theta$ should have a minus sign.
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
$endgroup$
From your first line to your second you did not square correctly. If you are squaring the equation you missed the cross term $4sin^2 theta cos theta$. If you took the $sin theta$ to the other side before squaring to avoid the cross term, when you brought it back the $sin^2 theta$ should have a minus sign.
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 15:08
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
add a comment |
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
$begingroup$
Hi Ross thanks a lot for your solution that's great, I've been getting those wrong all day. One question is that using your latter suggestion and coming to an answer of $3sin^2 - 4sin^4 = 0$ this gives a solution of $sqrt(3)/2$ which through arcsin gives 60 degrees, rather than 120, so still not sure where I am going wrong. Thanks a lot.
$endgroup$
– Joseph
Feb 1 at 15:47
1
1
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
The arcsine is defined in the range $[-90^circ,90^circ]$ to make it a function. You know you need some angle that has a sine of $frac {sqrt 3}2$. Both $60^circ$ and $120^circ$ satisfy that. You introduced $60^circ$ as an extraneous solution with your square. The usual thing of plugging back into the original equation will tell you which is correct. Here you need the cosine to be negative, which tells you that the solution has to be in the second quadrant.
$endgroup$
– Ross Millikan
Feb 1 at 15:51
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
$begingroup$
Thanks a lot for your help Ross, that was incredibly useful.
$endgroup$
– Joseph
Feb 15 at 18:13
add a comment |
$begingroup$
I'll start by graphing this function x-axis is $theta / pi $ which shows the function is zero at 5 points.
begin{align}
2 cdot sin(theta)cos(theta) + sin(theta) & = 0 \
sin(theta) cdot (2cdotcos(theta)+1) & = 0
end{align}
So either $sin(theta) = 0$ or $2cdotcos(theta)+1 = 0 Rightarrow cos(theta) =
-0.5$
Considering each of these cases then $theta$ is $-pi$, $-dfrac{2}{3}pi$, $0$, $dfrac{2}{3}pi$ or $pi$
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
$endgroup$
add a comment |
$begingroup$
I'll start by graphing this function x-axis is $theta / pi $ which shows the function is zero at 5 points.
begin{align}
2 cdot sin(theta)cos(theta) + sin(theta) & = 0 \
sin(theta) cdot (2cdotcos(theta)+1) & = 0
end{align}
So either $sin(theta) = 0$ or $2cdotcos(theta)+1 = 0 Rightarrow cos(theta) =
-0.5$
Considering each of these cases then $theta$ is $-pi$, $-dfrac{2}{3}pi$, $0$, $dfrac{2}{3}pi$ or $pi$
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
$endgroup$
add a comment |
$begingroup$
I'll start by graphing this function x-axis is $theta / pi $ which shows the function is zero at 5 points.
begin{align}
2 cdot sin(theta)cos(theta) + sin(theta) & = 0 \
sin(theta) cdot (2cdotcos(theta)+1) & = 0
end{align}
So either $sin(theta) = 0$ or $2cdotcos(theta)+1 = 0 Rightarrow cos(theta) =
-0.5$
Considering each of these cases then $theta$ is $-pi$, $-dfrac{2}{3}pi$, $0$, $dfrac{2}{3}pi$ or $pi$
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
$endgroup$
I'll start by graphing this function x-axis is $theta / pi $ which shows the function is zero at 5 points.
begin{align}
2 cdot sin(theta)cos(theta) + sin(theta) & = 0 \
sin(theta) cdot (2cdotcos(theta)+1) & = 0
end{align}
So either $sin(theta) = 0$ or $2cdotcos(theta)+1 = 0 Rightarrow cos(theta) =
-0.5$
Considering each of these cases then $theta$ is $-pi$, $-dfrac{2}{3}pi$, $0$, $dfrac{2}{3}pi$ or $pi$
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
answered Feb 1 at 16:16
Warren HillWarren Hill
2,7091022
2,7091022
add a comment |
add a comment |
$begingroup$
For $-pileqthetaleqpi$:
begin{align*}
2sin{theta}cos{theta}+sin{theta}&=0 \
sin{theta}big(2cos{theta}+1big)&=0 \
end{align*}
Thus, $sin{theta}=0$ or $2cos{theta}+1=0$ .
- If $sin{theta}=0$, then $thetain{-pi,0,pi}$ .
- If $2cos{theta}+1=0$, then $cos{theta}=-1/2<0$, then $thetain(-pi,-pi/2)cup(pi/2,pi)$ and thus $thetain{-2pi/3,2pi/3}$
The solution is $thetainbigg{-pi,-dfrac{2pi}{3},0,dfrac{2pi}{3},pibigg}$ .
answered Feb 1 at 15:14
El boritoEl borito
664216
$endgroup$
add a comment |
$begingroup$
For $-pileqthetaleqpi$:
begin{align*}
2sin{theta}cos{theta}+sin{theta}&=0 \
sin{theta}big(2cos{theta}+1big)&=0 \
end{align*}
Thus, $sin{theta}=0$ or $2cos{theta}+1=0$ .
- If $sin{theta}=0$, then $thetain{-pi,0,pi}$ .
- If $2cos{theta}+1=0$, then $cos{theta}=-1/2<0$, then $thetain(-pi,-pi/2)cup(pi/2,pi)$ and thus $thetain{-2pi/3,2pi/3}$
The solution is $thetainbigg{-pi,-dfrac{2pi}{3},0,dfrac{2pi}{3},pibigg}$ .
answered Feb 1 at 15:14
El boritoEl borito
664216
$endgroup$
add a comment |
$begingroup$
For $-pileqthetaleqpi$:
begin{align*}
2sin{theta}cos{theta}+sin{theta}&=0 \
sin{theta}big(2cos{theta}+1big)&=0 \
end{align*}
Thus, $sin{theta}=0$ or $2cos{theta}+1=0$ .
- If $sin{theta}=0$, then $thetain{-pi,0,pi}$ .
- If $2cos{theta}+1=0$, then $cos{theta}=-1/2<0$, then $thetain(-pi,-pi/2)cup(pi/2,pi)$ and thus $thetain{-2pi/3,2pi/3}$
The solution is $thetainbigg{-pi,-dfrac{2pi}{3},0,dfrac{2pi}{3},pibigg}$ .
answered Feb 1 at 15:14
El boritoEl borito
664216
$endgroup$
For $-pileqthetaleqpi$:
begin{align*}
2sin{theta}cos{theta}+sin{theta}&=0 \
sin{theta}big(2cos{theta}+1big)&=0 \
end{align*}
Thus, $sin{theta}=0$ or $2cos{theta}+1=0$ .
- If $sin{theta}=0$, then $thetain{-pi,0,pi}$ .
- If $2cos{theta}+1=0$, then $cos{theta}=-1/2<0$, then $thetain(-pi,-pi/2)cup(pi/2,pi)$ and thus $thetain{-2pi/3,2pi/3}$
The solution is $thetainbigg{-pi,-dfrac{2pi}{3},0,dfrac{2pi}{3},pibigg}$ .
answered Feb 1 at 15:14
El boritoEl borito
664216
edited Feb 2 at 5:06
answered Feb 1 at 15:14
El boritoEl borito
664216
answered Feb 1 at 15:14
El boritoEl borito
664216
answered Feb 1 at 15:14
El boritoEl borito
664216
664216
add a comment |
add a comment |
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1
$begingroup$
How did you square?
$endgroup$
– Math-fun
Feb 1 at 15:06