Mathematics - Trigonometric Equation Question with Solution | TestHub

Given that

${\sin }^{5}x-{\cos }^{5}x=\frac{\sin x-\cos x}{\sin x\cos x}$

$$\begin{array}{rl}\sin x\cos x\left[\frac{{\sin }^{5}x-{\cos }^{5}x}{\sin x-\cos x}\right]& =1\\ \sin x\cos x\left\{{\sin }^{4}x+{\sin }^{3}x\cos x+{\sin }^{2}x{\cos }^{2}x+\sin x\cos x({\sin }^{2}x+{\cos }^{2}x)\right\}& =1\\ \sin x\cos x\left\{({\sin }^{2}x+{\cos }^{2}x{)}^2-{\sin }^{2}x{\cos }^{2}x+\sin x\cos x\right\}& =1\\ \Rightarrow \frac{1}{2}\sin 2x\left[1-\frac{1}{4}{\sin }^{2}2x+\frac{1}{2}\sin 2x\right]& =1\\ \sin 2x({\sin }^{2}2x-2\sin 2x-4)& =-8\\ {\sin }^{3}2x-2{\sin }^{2}2x-4\sin 2x+8& =0\\ (\sin 2x-2{)}^2(\sin 2x+2)& =0\\ \Rightarrow \sin 2x& =\pm 2\end{array}$$

which is impossible