Answer: \(\displaystyle x=\frac{11\pi}{24}\) or \(\displaystyle x=\frac{23\pi}{24}\).

Answer: \(\displaystyle x=\frac{11\pi}{24}\) or \(\displaystyle x=\frac{23\pi}{24}\).

Let

\(u=2x+\frac{\pi}{4}.\)

The equation becomes

\(3\sin u=\sqrt3\cos u.\)

Since \(\cos u\neq0\) for the solutions of this equation, divide by \(\cos u\):

\(3\tan u=\sqrt3.\)

Therefore

\(\tan u=\frac{1}{\sqrt3}.\)

As \(0\leq x\leq\pi\),

\(\frac{\pi}{4}\leq u\leq\frac{9\pi}{4}.\)

In this interval,

\(u=\frac{7\pi}{6}\quad\text{or}\quad u=\frac{13\pi}{6}.\)

Hence

\(2x+\frac{\pi}{4}=\frac{7\pi}{6}\)

or

\(2x+\frac{\pi}{4}=\frac{13\pi}{6}.\)

Solving gives

\(x=\frac{11\pi}{24}\quad\text{or}\quad x=\frac{23\pi}{24}.\)