Trigonometry - Double angle

Question

I have tried to solve this problem, but everything that I try does not work. Please help me solve this equation:

$$\cos {6}x + 2 = 5\sin {3}x$$

Thanks :)

Answer 1

Hint: $$\cos 2 \theta = 1- 2 \sin^2 \theta$$

Answer 2

Let's pursue the path shown by @Henry. $\cos{6x}=1-2\sin^2{3x}$. So our equation becomes

$$2\sin^2{3x}+5\sin{3x}-3=0$$

Set $U=\sin{3x}$ and let's solve the quadratic $2U^2+5U-3=0$. It has two solutions $U_1=-3$ and $U_2={1\over 2}$. Only $U_2$ can be a sine because $|U_1|\gt 1$. So our problem is about solving

$$\sin{3x}={1\over 2}=\sin{{\pi\over 6}}$$

And we get $x={\pi\over 18}+{2k\pi\over 3}$ with $k=0,\,1,\,2$