Help PLz :)

$2\cos\theta+\sin(2\theta)=0$

Let's use the double-angle formula for sin which says  $\sin(2\theta)=2\sin\theta\cos\theta$

$2\cos\theta+2\sin\theta\cos\theta=0$

Now we can factor  2 cos θ  out of both terms on the left side of the equation.

$2\cos\theta(1+\sin\theta)=0$      (Notice if we distribute  2 cos θ  we get the previous expression)

Set each factor equal to  0  and solve for  θ:

$\begin{array}{ccc} 2\cos\theta=0&\text{or}&1+\sin\theta=0\\~\\ \cos\theta=0&\text{or}&\sin\theta=-1\\~\\ \theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\dots&\text{or}&\theta=\frac{3\pi}{2},\frac{7\pi}{2},\frac{11\pi}{2},\frac{15\pi}{2},\dots\\~\\ \theta=90^\circ,270^\circ,450^\circ,630^\circ,\dots&\text{or}&\theta=270^\circ,630^\circ,990^\circ,1350^\circ\dots \end{array}$

The solutions in the interval  [0°, 360°)  are:   90°, 270°