Answer: \(\alpha=\frac{7\pi}{24}\) or \(\alpha=\frac{23\pi}{24}\).

(a) Given

\(2\cos x=3\tan x,\)

write \(\tan x=\dfrac{\sin x}{\cos x}\):

\(2\cos x=3\frac{\sin x}{\cos x}.\)

Multiplying by \(\cos x\),

\(2\cos^2x=3\sin x.\)

Using \(\cos^2x=1-\sin^2x\),

\(2(1-\sin^2x)=3\sin x.\)

Rearranging,

\(2\sin^2x+3\sin x-2=0.\)

(b) Let

\(X=2\alpha+\frac{\pi}{4}.\)

From part (a),

\(2\sin^2X+3\sin X-2=0.\)

Factorising,

\((2\sin X-1)(\sin X+2)=0.\)

Since \(\sin X=-2\) is impossible,

\(\sin X=\frac12.\)

Now \(0\lt \alpha\lt \pi\), so

\(\frac{\pi}{4}\lt X\lt \frac{9\pi}{4}.\)

In this interval,

\(X=\frac{5\pi}{6}\quad\text{or}\quad X=\frac{13\pi}{6}.\)

Therefore

\(2\alpha+\frac{\pi}{4}=\frac{5\pi}{6}\)

or

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

Solving gives

\(\alpha=\frac{7\pi}{24}\)

or

\(\alpha=\frac{23\pi}{24}.\)