Answer: (a)(ii) \(\theta=2.64,\ 9.93\) radians; (b) \(y=172^\circ,\ 352^\circ\).
(a)(i) Use \(\tan x=\dfrac{\sin x}{\cos x}\):
\(\sin x\tan x+\cos x =\sin x\left(\frac{\sin x}{\cos x}\right)+\cos x.\)
So
\(\sin x\tan x+\cos x =\frac{\sin^2x}{\cos x}+\frac{\cos^2x}{\cos x} =\frac{\sin^2x+\cos^2x}{\cos x}.\)
Since \(\sin^2x+\cos^2x=1\),
\(\sin x\tan x+\cos x=\frac1{\cos x}=\operatorname{sec}x.\)
(a)(ii) Put \(x=\dfrac{\theta}{2}\). From part (a)(i),
\(\sin\frac{\theta}{2}\tan\frac{\theta}{2}+\cos\frac{\theta}{2} =\operatorname{sec}\frac{\theta}{2}.\)
The equation becomes
\(\operatorname{sec}\frac{\theta}{2}=4.\)
Hence
\(\cos\frac{\theta}{2}=\frac14.\)
Let \(u=\dfrac{\theta}{2}\). Since \(0\leqslant\theta\leqslant4\pi\),
\(0\leqslant u\leqslant2\pi.\)
Solving \(\cos u=\frac14\) gives
\(u=1.3181\ldots\quad\text{or}\quad u=4.9651\ldots.\)
Therefore
\(\theta=2u=2.6362\ldots\quad\text{or}\quad 9.9301\ldots.\)
So
\(\theta=2.64,\ 9.93\)
to 3 significant figures.
(b) Since \(\operatorname{cot}(y+38^\circ)=\sqrt3\),
\(\tan(y+38^\circ)=\frac1{\sqrt3}.\)
Therefore
\(y+38^\circ=30^\circ+180^\circ k.\)
So
\(y=-8^\circ+180^\circ k.\)
In the interval \(0^\circ\leqslant y\leqslant360^\circ\), this gives
\(y=172^\circ,\ 352^\circ.\)
