Answer: \(x=57.7^\circ,\ 122.3^\circ,\ 237.7^\circ,\ 302.3^\circ\).
(a) Start with the left side:
\(\frac1{\operatorname{cosec}x-1}+\frac1{\operatorname{cosec}x+1}.\)
Use a common denominator:
\(\frac{\operatorname{cosec}x+1+\operatorname{cosec}x-1} {(\operatorname{cosec}x-1)(\operatorname{cosec}x+1)} =\frac{2\operatorname{cosec}x}{\operatorname{cosec}^2x-1}.\)
Since \(\operatorname{cosec}^2x-1=\operatorname{cot}^2x\),
\(\frac{2\operatorname{cosec}x}{\operatorname{cosec}^2x-1} =\frac{2\operatorname{cosec}x}{\operatorname{cot}^2x}.\)
Now use \(\operatorname{cosec}x=\frac1{\sin x}\) and \(\operatorname{cot}^2x=\frac{\cos^2x}{\sin^2x}\):
\(\frac{2\operatorname{cosec}x}{\operatorname{cot}^2x} =\frac2{\sin x}\cdot\frac{\sin^2x}{\cos^2x} =\frac{2\sin x}{\cos^2x}.\)
Also
\(2\tan x\operatorname{sec}x =2\cdot\frac{\sin x}{\cos x}\cdot\frac1{\cos x} =\frac{2\sin x}{\cos^2x}.\)
Therefore
\(\frac1{\operatorname{cosec}x-1}+\frac1{\operatorname{cosec}x+1} =2\tan x\operatorname{sec}x.\)
(b) Using the identity from part (a), the equation becomes
\(2\tan x\operatorname{sec}x=5\operatorname{cosec}x.\)
In sine and cosine form,
\(2\cdot\frac{\sin x}{\cos x}\cdot\frac1{\cos x} =5\cdot\frac1{\sin x}.\)
So
\(\frac{2\sin x}{\cos^2x}=\frac5{\sin x}.\)
Cross-multiplying gives
\(2\sin^2x=5\cos^2x.\)
Divide by \(\cos^2x\):
\(2\tan^2x=5.\)
Hence
\(\tan x=\pm\sqrt{\frac52}.\)
The acute angle is
\(\tan^{-1}\sqrt{\frac52}=57.688\ldots^\circ.\)
For \(0^\circ\lt x\lt 360^\circ\),
\(x=57.7^\circ,\ 122.3^\circ,\ 237.7^\circ,\ 302.3^\circ\)
to 1 decimal place.
