Answer: \(k=3\); \(x=\pm0.955,\ \pm2.19\); \(\frac{\mathrm dy}{\mathrm dx}=\frac{2\sqrt{x}-1}{2\sqrt{x}\cos^2(x-\sqrt{x})}\); integral \(=2\tan(x-\sqrt{x})+C\).

We start with the main method. Use the product, chain and standard trigonometric derivative rules carefully.

(a)(i) Differentiate

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

Then

\(\frac{\mathrm dy}{\mathrm dx}=6\sin x\cos x-\sin x.\)

Now

\(\cot x\frac{\mathrm dy}{\mathrm dx} =\frac{\cos x}{\sin x}(6\sin x\cos x-\sin x) =6\cos^2x-\cos x.\)

Therefore

\(y+\cot x\frac{\mathrm dy}{\mathrm dx} =3\sin^2x+\cos x+6\cos^2x-\cos x.\)

So

\(y+\cot x\frac{\mathrm dy}{\mathrm dx}=3\sin^2x+6\cos^2x.\)

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

\(3\sin^2x+6\cos^2x=3(1-\cos^2x)+6\cos^2x=3(1+\cos^2x).\)

Hence \(k=3\).

(a)(ii) Solve

\(3(1+\cos^2x)=4.\)

This gives

\(\cos^2x=\frac13.\)

So \(\cos x=\pm\frac1{\sqrt3}\). For \(-\pi\leq x\leq\pi\), the solutions are

\(x=\pm0.955,\quad \pm2.19\) radians.

(b)(i) For \(y=\tan(x-\sqrt{x})\), use the chain rule:

\(\frac{\mathrm dy}{\mathrm dx}=\sec^2(x-\sqrt{x})\left(1-\frac1{2\sqrt{x}}\right).\)

Thus

\(\frac{\mathrm dy}{\mathrm dx} =\frac{2\sqrt{x}-1}{2\sqrt{x}\cos^2(x-\sqrt{x})}.\)

(b)(ii) The integrand is twice the derivative found in part (b)(i):

\(\frac{2\sqrt{x}-1}{\sqrt{x}\cos^2(x-\sqrt{x})} =2\cdot\frac{2\sqrt{x}-1}{2\sqrt{x}\cos^2(x-\sqrt{x})}.\)

Therefore

\(\int\frac{2\sqrt{x}-1}{\sqrt{x}\cos^2(x-\sqrt{x})}\,\mathrm dx =2\tan(x-\sqrt{x})+C.\)

This completes the solution and gives the required result.