Answer: (a) \(\dfrac{dy}{dx}=\dfrac{\cos x}{1+\sin x}\); (b) \(\dfrac1{\sqrt3}\); (c) \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{5\pi}{6}\).

(a) Differentiate \(y=\ln(1+\sin x)\) using the chain rule.

If \(u=1+\sin x\), then \(\dfrac{du}{dx}=\cos x\), so

\(\frac{dy}{dx}=\frac{1}{1+\sin x}\cdot\cos x.\)

Hence

\(\frac{dy}{dx}=\frac{\cos x}{1+\sin x}.\)

(b) When \(x=\dfrac{\pi}{6}\),

\(\cos\frac{\pi}{6}=\frac{\sqrt3}{2},\qquad \sin\frac{\pi}{6}=\frac12.\)

Therefore

\(\frac{dy}{dx}=\frac{\frac{\sqrt3}{2}}{1+\frac12} =\frac{\frac{\sqrt3}{2}}{\frac32} =\frac{\sqrt3}{3}.\)

Since \(\dfrac{\sqrt3}{3}=\dfrac1{\sqrt3}\), the answer is

\(\frac1{\sqrt3}.\)

(c) Solve

\(\frac{\cos x}{1+\sin x}=\tan x.\)

Since \(\tan x=\dfrac{\sin x}{\cos x}\),

\(\frac{\cos x}{1+\sin x}=\frac{\sin x}{\cos x}.\)

Multiply by \((1+\sin x)\cos x\):

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

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

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

So

\(2\sin^2x+\sin x-1=0.\)

Factorising,

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

In the interval \(0\lt x\lt\pi\), the solutions come from

\(\sin x=\frac12.\)

Therefore

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