# Thread: trig equation solve for theta

1. ## trig equation solve for theta

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

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

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

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

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

can you solve now?

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

for values of $\displaystyle \theta$ 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
$\displaystyle cos3\theta = cos^{3}\theta - 3cos\theta sin^{2}\theta$

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

can you solve now?
would it be like this

$\displaystyle cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ 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
$\displaystyle \cos A = \cos B$
has the general solution
$\displaystyle A = 2n\pi \pm B$
So, with the equation
$\displaystyle 3\theta = 2n\pi \pm(\pi-\theta)$
take the $\displaystyle +$ sign and $\displaystyle -$ sign one at a time; solve for $\displaystyle \theta$, and then plug in various values of $\displaystyle n\; (0, \pm1,\pm2,...)$ until you have all the values of $\displaystyle \theta$ in the range you want.

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

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

$\displaystyle 4\theta={\pi}$

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

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

there are three more solutions until the solutions repeat themselves.

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

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

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

7. Originally Posted by sigma1
would it be like this

$\displaystyle cos^3\theta - 3cos\theta sin^2\theta + cos\theta =0$ 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 $\displaystyle sin^2(x)+ cos^2(x)= 1$, $\displaystyle sin^2(x)= 1- cos^2(x)$ so this becomes
$\displaystyle cos^3(\theta)- 3cos(\theta)(1- cos^2(\theta)+ cos(\theta)= 0$

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

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

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

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

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