## Trigonometry 7th Edition

$$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$
Given $$\cos 2x \cos x - \sin 2x \sin x=\frac{ \sqrt{2}}{2},\ \ 0\leq x<2\pi$$ Then using $\cos(A+B)= \cos A \cos B - \sin A \sin B$ , we get \begin{align*} \cos 2x \cos x - \sin 2x \sin x&=\frac{ \sqrt{2}}{2}\\ \cos(3x)&=\frac{ \sqrt{2}}{2}\\ \end{align*} First we find all possible solutions for $x$ $$3x=\frac{\pi}{4}+2k\pi \ \Rightarrow x= \frac{\pi}{12}+ \frac{2k\pi}{3}$$ To find those solutions that lie in the interval $0\leq x<2\pi$ we let k take on values of 0, 1, and 2; we get $$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$