Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$

Question

Find all solutions for $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$ if $0\leq x< \pi$

Can you verify my work? Thanks!

$$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$$

$$\cos(2x+x)=\frac{1}{\sqrt{2}}$$

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

$$\cos^{-1}(\frac{1}{\sqrt{2}})$$

Reference angle: $\frac{\pi}{4}$

$$3x=\frac{\pi}{4}+2k\pi$$

$$x=\frac{\pi}{12}+\frac{2k\pi}{3}$$

$$3x=\frac{7\pi}{4}+2k\pi$$

$$x=\frac{7\pi}{12}+\frac{2k\pi}{3}$$

$$x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12}, \frac{7\pi}{12},\frac{5\pi}{4},\frac{23\pi}{12} $$

Answer 1

Right idea, but after reducing, you'll get cos(3x) = 0 (reference circle), so you'll be left with x=π/6, x=π/2 and x=5π/6 after simplifying.

Realize that 5π/4 and 23π/12 are >π, all larger than π will not be included with your solution.

EDIT: SEE COMMENT BELOW.

Answer 2

The solution $x=\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angles $\frac{\pi}{12}$ and $\frac{9\pi}{12}=\frac{3\pi}{4}$.

And the solution $x=-\frac{\pi}{12} + \frac{2k\pi}{3}$ gives, in $[0, \pi[$, the angle $\frac{7\pi}{12}$.

And that's all.