Answer: \(x=39.2^\circ,\ 140.8^\circ,\ 219.2^\circ,\ 320.8^\circ\).
(a) Start with the left side:
\(\frac1{\operatorname{sec}x-1}+\frac1{\operatorname{sec}x+1}.\)
Use a common denominator:
\(\frac{\operatorname{sec}x+1+\operatorname{sec}x-1} {(\operatorname{sec}x-1)(\operatorname{sec}x+1)} =\frac{2\operatorname{sec}x}{\operatorname{sec}^2x-1}.\)
Since \(\operatorname{sec}^2x-1=\tan^2x\),
\(\frac{2\operatorname{sec}x}{\operatorname{sec}^2x-1} =\frac{2\operatorname{sec}x}{\tan^2x}.\)
Now use \(\operatorname{sec}x=\frac1{\cos x}\) and \(\tan^2x=\frac{\sin^2x}{\cos^2x}\):
\(\frac{2\operatorname{sec}x}{\tan^2x} =\frac{2}{\cos x}\cdot\frac{\cos^2x}{\sin^2x} =\frac{2\cos x}{\sin^2x}.\)
Also
\(2\operatorname{cot} x\operatorname{cosec}x =2\cdot\frac{\cos x}{\sin x}\cdot\frac1{\sin x} =\frac{2\cos x}{\sin^2x}.\)
Therefore
\(\frac1{\operatorname{sec}x-1}+\frac1{\operatorname{sec}x+1} =2\operatorname{cot} x\operatorname{cosec}x.\)
(b) Using the identity from part (a), the equation becomes
\(2\operatorname{cot} x\operatorname{cosec}x=3\operatorname{sec}x.\)
In terms of sine and cosine, this is
\(2\cdot\frac{\cos x}{\sin x}\cdot\frac1{\sin x} =\frac3{\cos x}.\)
So
\(\frac{2\cos x}{\sin^2x}=\frac3{\cos x}.\)
Cross-multiplying gives
\(2\cos^2x=3\sin^2x.\)
Divide by \(\cos^2x\):
\(2=3\tan^2x.\)
Hence
\(\tan^2x=\frac23.\)
So
\(\tan x=\pm\sqrt{\frac23}.\)
The acute angle is
\(\tan^{-1}\sqrt{\frac23}=39.2315\ldots^\circ.\)
For \(0^\circ\lt x\lt 360^\circ\),
\(x=39.2^\circ,\ 140.8^\circ,\ 219.2^\circ,\ 320.8^\circ\)
to 1 decimal place.
