Solving $\sin \frac12x =-3 \cos 2x$

Question

I have been trying to solve this question:

$$\sin \frac12x =-3 \cos 2x$$

However, I do not even know where to begin. I have graphed it, so I know what x must be and that there are 4 solutions, but I just do not know the equations.

What should I do to solve it?

Answer 1

$$\sin\left(\frac{x}{2}\right) = \sqrt[]{\frac{1-\cos(x)}{2}} = -3\cos(2x)$$ $$\frac{1-\cos(x)}{2} = 9\cos^2(2x) \rightarrow 0= 18\cos^2(2x)+\cos(x)-1$$

From here it is a lot of bashing/algebra knowing that $\cos(2x) = \sin^2(x)-\cos^2(x)$

Answer 2

Let $\sin\dfrac x2=a,\cos x=1-2a^2$

$$\implies a=-3\left(2(1-2a^2)^2-1\right)$$

$$24a^4-24a^2+a+3=0$$