## Algebra 2 (1st Edition)

$x=\dfrac{3 \pi}{2}$
Here, we have $\cos (x+\dfrac{\pi}{6})-1= \cos (x -\dfrac{\pi}{6})$ Re-write as: $\cos (x+\dfrac{\pi}{6})- \cos (x -\dfrac{\pi}{6})=1$ We know that $\cos a-\cos b=-2 \sin (\dfrac{a-b}{2})(\dfrac{a+b}{2})$ Thus, we have $-2 \sin (\dfrac{(x+\dfrac{\pi}{6})-(x-\dfrac{\pi}{6})}{2})(\dfrac{(x+\dfrac{\pi}{6})+(x-\dfrac{\pi}{6})}{2})=1$ $-2 \sin (\pi/6) \sin x=1$ $-2 \times (1/2) \sin x=1 \implies \sin x=-1$ Thus, in the interval $0 \leq x \leq 2 \pi$, we have only $x=\dfrac{3 \pi}{2}$