<s>[INST] Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$ 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: $\frac{\pi}{4}$
$$3x=\frac{\pi}{4}+2k\pi$$   
$$x=\frac{\pi}{12}+\frac{2k\pi}{3}$$
$$3x=\frac{7\pi}{4}+2k\pi$$
$$x=\frac{7\pi}{12}+\frac{2k\pi}{3}$$
$$x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12}, \frac{7\pi}{12},\frac{5\pi}{4},\frac{23\pi}{12} $$
 [/INST] 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.
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