A trigonometric problem

Question

$$2 (\cos{⁡2x} )(\sin ⁡2x-1)=\sqrt{3} \cos{⁡4}x$$ I have tried to reduce this equation and I got this: $\cos(4x+\pi/6)=-\cos2x$ but I don't know how to do next

Answer 1

We know that: $$\cos(a)=\cos(b) \leftrightarrow a=b+\pi+2k\pi \vee a+b=\pi+2k\pi$$ Here $a=4x+\frac{\pi}{6}$ and $b=-2x$ because we know that $-\cos(2x)=\cos(-2x)$ In the first case, we have: $$4x+\frac{\pi}{6}=\pi-2x+2k\pi \leftrightarrow x=\frac{5\pi}{36}+\frac{k\pi}{3}$$ In the second case, we obtain: $$4x+\frac{\pi}{6}-2x=\pi+2k\pi \leftrightarrow x=\frac{5\pi}{12}+k\pi$$

Answer 2

Hint:

You can write $\;-\cos2x=\cos (2x+\pi)$ and use that $$ \cos\theta=\cos\varphi\iff \theta\equiv \pm\varphi \pmod{2\pi}. $$