# How do you find all solutions for 4sinthetacostheta=sqrt(3) for the interval [0,2pi]?

Oct 24, 2014

4 $\sin \theta \cos \theta$ = root3
2 $\sin 2 \theta$ = root3
$\sin 2 \theta$= (root3) /2

$2 \theta$ = $n \pi$ + ${\left(- 1\right)}^{n} \frac{\pi}{3}$

$\theta$ = $n \frac{\pi}{2}$ + ${\left(- 1\right)}^{n} \frac{\pi}{6}$

For n=0, $\theta$= $\frac{\pi}{6}$

For n=1, $\theta$= $\frac{\pi}{3}$

For n=2, $\theta$= $7 \frac{\pi}{6}$

For n=3, $\theta$= $3 \frac{\pi}{2}$ - $\frac{\pi}{6}$

For n=4, $\theta$ becomes greater than $2 \pi$.
THus the values of $\theta$ for n= 0, 1, 2, 3 are the solutions.