# Thread: trig equation solve for theta

1. ## trig equation solve for theta

hello how would i solve for $cos3\theta+cos\theta=0$

for values of $\theta$ between 0 and 360 degrees

2. Originally Posted by sigma1
hello how would i solve for $cos3\theta+cos\theta=0$

for values of $\theta$ between 0 and 360 degrees
$cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta$

convert $\sin\theta$ in terms of $cos\theta$

can you solve now?

3. Originally Posted by sigma1
hello how would i solve for $cos3\theta+cos\theta=0$

for values of $\theta$ between 0 and 360 degrees
You can rewrite the equation as
cos(3θ) = -cos(θ) = cos(π - θ)
Therefore the general solution is
3θ = 2nπ + or - (π - θ)
Now solve for θ.

4. Originally Posted by harish21
$cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta$

convert $\sin\theta$ in terms of $cos\theta$

can you solve now?
would it be like this

$cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ if so where do i go from here? can u help me solve this?

5. Hello sigma1

sa-ri-ga-ma's solution is the easier method. He has used the fact that
$\cos A = \cos B$
has the general solution
$A = 2n\pi \pm B$
So, with the equation
$3\theta = 2n\pi \pm(\pi-\theta)$
take the $+$ sign and $-$ sign one at a time; solve for $\theta$, and then plug in various values of $n\; (0, \pm1,\pm2,...)$ until you have all the values of $\theta$ in the range you want.

6. Some graphs for sa-ri-ga-ma's solution.

$3\theta={\pi}-\theta$

$4\theta={\pi}$

$3\theta=2n{\pi}+({\pi}-\theta)$

$4\theta=2n{\pi}+{\pi}$

there are three more solutions until the solutions repeat themselves.

$3\theta=2n{\pi}-({\pi}-\theta)$

$3\theta=2n{\pi}-{\pi}+\theta$

$2\theta=2n{\pi}-{\pi}$

7. Originally Posted by sigma1
would it be like this

$cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ if so where do i go from here? can u help me solve this?
harish21 said "convert sine in terms of cosine" and you have not done that yet. Since $sin^2(x)+ cos^2(x)= 1$, $sin^2(x)= 1- cos^2(x)$ so this becomes
$cos^3(\theta)- 3cos(\theta)(1- cos^2(\theta)+ cos(\theta)= 0$

$cos^3(\theta)- 3 cos(\theta)+ 3 cos^3(\theta)+ cos(\theta)= 0$

$4 cos^3(\theta)- 2 cos(\theta)= 0$

If you like, you can replace $cos(\theta)$ with "x" so you have the equation
$4x^3- 2x= x(4x^2- 2)= 0$

Either x= 0 or $4x^2- 2= 0$ which leads to $x^2= \frac{1}{2}$ so that $x= \pm\frac{1}{\sqrt{2}}= \pm\frac{\sqrt{2}}{2}$.

Now you need to solve $y= cos(\theta)= 0$, $cos(\theta)= \frac{\sqrt{2}}{2}$, and $cos(\theta)= -\frac{\sqrt{2}}{2}$.