Answer: \(m=6\); \(x=35.3^\circ,\ 144.7^\circ,\ 215.3^\circ,\ 324.7^\circ\).

(a) Since

\(y=3\tan^2x,\)

differentiate using the chain rule:

Therefore

\(\frac{dy}{dx}=6\tan x\operatorname{sec}^2x,\)

so

\(m=6.\)

(b) Set the two expressions for \(\frac{dy}{dx}\) equal:

Divide by \(3\):

Using \(\tan x=\frac{\sin x}{\cos x}\), \(\operatorname{sec}x=\frac1{\cos x}\) and \(\operatorname{cosec}x=\frac1{\sin x}\),

Multiplying by \(\sin x\cos^3x\),

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

Divide by \(\cos^2x\):

\(2\tan^2x=1.\)

Hence

\(\tan x=\pm\frac1{\sqrt2}.\)

The acute angle is

For \(0^\circ\lt x\lt 360^\circ\), the solutions are

\(x=35.3^\circ,\ 144.7^\circ,\ 215.3^\circ,\ 324.7^\circ\)

to 1 decimal place.