hello how would i solve for $\displaystyle cos3\theta+cos\theta=0$
for values of $\displaystyle \theta$ between 0 and 360 degrees
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.
Grandad
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.
$\displaystyle 3\theta=2n{\pi}-({\pi}-\theta)$
$\displaystyle 3\theta=2n{\pi}-{\pi}+\theta$
$\displaystyle 2\theta=2n{\pi}-{\pi}$
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 $\displaystyle cos(\theta)= -\frac{\sqrt{2}}{2}$.