Find coordinates of intersection

Question

The question says "The line with equation $y = - \sqrt{3}$ intersects the graph at points A and B, find coordinates of point B." I worked out that the graph formula is $y = 2\cos(2x)$ and I think I'm supposed to use simultaneous equations but i wasn't sure how to solve $2\cos(2x) = - \sqrt{3}$

Answer 1

$$2\cos(2x) = - \sqrt{3}\tag{1}$$

Recall the identity $$\cos(2x) = 1-2\sin^2(x)$$ Then $(1)$ becomes $$2-4\sin^2(x)=-\sqrt{3}$$ $$\implies x=\pm \sin^{-1}\left(\sqrt{\cfrac{2+\sqrt{3}}{4}}\right)=\pm 75^{\circ}$$

To verify. From $(1)$ $$\cos(2x)=\frac{- \sqrt{3}}{2}\tag{2}$$

Insertion of $75^{\circ}$ into $(2)$ yields $$\cos(150)=\frac{- \sqrt{3}}{2}$$ and since $\cos (2x)$ is even such that $\cos(2x)=\cos(-2x)$ $$\cos(-150)=\frac{- \sqrt{3}}{2}$$

Answer 2

Hint:

Do you know an angle with cosine equal to $\dfrac{\sqrt 3}2$?

When you've written the equation in the form $\;\cos 2x=\cos\alpha$, the general solution is $$2x\equiv \pm\alpha\mod2\pi\iff x\equiv\pm\frac\alpha2\mod\pi.$$