Question

question 13 (5 points)
which of the following are the solutions of \\( \cos \left( \frac { \pi } { 6 } + x \
ight) + \sin \left( \frac { \pi } { 3 } + x \
ight) = 0 \\) on the interval \\( 0,2 \pi )? \\)
0
\\( 0, \pi \\)
\\( \frac { \pi } { 2 }, \frac { 3 \pi } { 2 } \\)
\\( \frac { \pi } { 2 } \\)

Explanation:

Step1: Use trigonometric identities

We know that \(\sin(A + B)=\sin A\cos B+\cos A\sin B\) and \(\cos(A + B)=\cos A\cos B-\sin A\sin B\).
First, \(\cos(\frac{\pi}{6}+x)=\cos\frac{\pi}{6}\cos x-\sin\frac{\pi}{6}\sin x=\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x\)
and \(\sin(\frac{\pi}{3}+x)=\sin\frac{\pi}{3}\cos x+\cos\frac{\pi}{3}\sin x=\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\)
Then \(\cos(\frac{\pi}{6}+x)+\sin(\frac{\pi}{3}+x)=(\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x)+(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)\)
Simplify the above - expression: \(\cos(\frac{\pi}{6}+x)+\sin(\frac{\pi}{3}+x)=\sqrt{3}\cos x\)

Step2: Solve the equation

Set \(\sqrt{3}\cos x = 0\)
We know that \(\cos x = 0\) when \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\)
On the interval \([0,2\pi]\), when \(k = 0\), \(x=\frac{\pi}{2}\); when \(k = 1\), \(x=\frac{3\pi}{2}\)

Answer:

\(\frac{\pi}{2},\frac{3\pi}{2}\)