Question

Solve: $1-\sin 2x=\cos x-\sin x$

Answer 1

Trigonometric equation solution:

$1-\sin 2x=\cos x-\sin x$

$$\begin{gathered} {(\cos x-\sin x)}^2=\cos x-\sin x \\ \Rightarrow \cos x-\sin x=1or\cos x-\sin x=0, \\ \Rightarrow \tan x=1 \\ \Rightarrow x=n\pi +\frac{\pi }{4} \\ \text{Now,} \\ \cos x-\sin x=1 \\ \Rightarrow \cos x-1=\sin x \\ \Rightarrow -2{\sin }^2\frac{x}{2}=2\sin \frac{x}{2}\cos \frac{x}{2} \\ \Rightarrow \sin \frac{x}{2}=0or\sin \frac{x}{2}=-\cos \frac{x}{2} \\ \Rightarrow \frac{x}{2}=n\mathrm{\pi }or\tan \frac{x}{2}=-1 \\ \Rightarrow x=2n\mathrm{\pi }or\frac{x}{2}=n\mathrm{\pi }–\frac{\mathrm{\pi }}{4} \\ \Rightarrow x=2n\mathrm{\pi }orx=2n\mathrm{\pi }–\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, general solutions are $x=2n\mathrm{\pi }orx=2n\mathrm{\pi }–\frac{\mathrm{\pi }}{2}orx=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$