# How do you find all solutions for sin 2x = cos x for the interval [0,2pi]?

Apr 28, 2018

The solutions are $S = \left\{\frac{1}{2} \pi , \frac{3}{2} \pi , \frac{1}{6} \pi , \frac{5}{6} \pi\right\}$

#### Explanation:

We need

$\sin 2 x = 2 \sin x \cos x$

Therefore,

$\sin 2 x = \cos x$

$\sin 2 x - \cos x = 0$

$2 \sin x \cos x - \cos x = 0$

$\cos x \left(2 \sin x - 1\right) = 0$

So,

$\left\{\begin{matrix}\cos x = 0 \\ 2 \sin x - 1 = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\cos x = 0 \\ \sin x = \frac{1}{2}\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = \frac{\pi}{2} & \frac{3}{2} \pi \\ x = \frac{1}{6} \pi & \frac{5}{6} \pi\end{matrix}\right.$ $\forall x \in \left[0 , 2 \pi\right]$

The solutions are $S = \left\{\frac{1}{2} \pi , \frac{3}{2} \pi , \frac{1}{6} \pi , \frac{5}{6} \pi\right\}$

graph{sin(2x)-cosx [-1.622, 9.475, -2.51, 3.04]}