Mathematics - Trigonometric Equation Question with Solution | TestHub

${\sin }^{2}x-{\cos }^{2}x=\sin x+\cos x$.

$-({\cos }^{2}x-{\sin }^{2}x)=\sin x+\cos x$

$-\cos (2x)=\sin x+\cos x$.

Using the identity $\cos (2x)={\cos }^{2}x-{\sin }^{2}x=(\cos x-\sin x)(\cos x+\sin x)$, $-(\cos x-\sin x)(\cos x+\sin x)=\sin x+\cos x$$(\sin x+\cos x)+(\cos x-\sin x)(\sin x+\cos x)=0$.

Factoring out $(\sin x+\cos x)$, we have $(\sin x+\cos x)[1+(\cos x-\sin x)]=0$. This gives two cases: Case 1: $\sin x+\cos x=0⟹\tan x=-1$. In $[0,4\pi ]$,

the solutions are $x=\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4},\frac{15\pi }{4}$. (4 solutions).

Case 2: $1+\cos x-\sin x=0⟹\sin x-\cos x=1$.

This can be written as $\sqrt{2}\sin (x-\frac{\pi }{4})=1$,

so $\sin (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$.

In $[0,4\pi ]$, $x-\frac{\pi }{4}$ ranges from $-\frac{\pi }{4}$ to $4\pi -\frac{\pi }{4}=\frac{15\pi }{4}$.

The general solutions for $\sin \theta =\frac{1}{\sqrt{2}}$ are $\theta =n\pi +(-1{)}^n\frac{\pi }{4}$. So $x-\frac{\pi }{4}=\frac{\pi }{4},\frac{3\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4}$.

This gives $x=\frac{\pi }{2},\pi ,\frac{5\pi }{2},3\pi$. (4 solutions).

No solutions from Case 1 overlap with solutions from Case 2 (e.g., $\tan (\pi /2)$ is undefined, $\tan (\pi )=0\ne -1$).

Therefore, the total number of solutions is $4+4=8$.