Mathematics - Trigonometric Equation Question with Solution | TestHub

The given equation is ${\sin }^{2}x-{\cos }^{2}x=\sin x-\cos x$. This can be rewritten as $-({\cos }^{2}x-{\sin }^{2}x)=\sin x-\cos x$, which simplifies to $-\cos 2x=\sin x-\cos x$. Factoring, we get$(\sin x-\cos x)(\sin x+\cos x)=\sin x-\cos x$. Rearranging, $(\sin x-\cos x)(\sin x+\cos x-1)=0$. Case 1:$\sin x-\cos x=0⟹\tan x=1$. Solutions are $x=n\pi +\frac{\pi }{4}$. In $[0,2\pi )$, these are $\frac{\pi }{4},\frac{5\pi }{4}$. Case 2:$\sin x+\cos x-1=0⟹\sqrt{2}\sin (x+\frac{\pi }{4})=1⟹\sin (x+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$. Solutions are $x+\frac{\pi }{4}=k\pi +(-1{)}^k\frac{\pi }{4}$. If$k$ is even, $x+\frac{\pi }{4}=2m\pi +\frac{\pi }{4}⟹x=2m\pi$. In $[0,2\pi )$, this is $x=0$. If$k$ is odd, $x+\frac{\pi }{4}=(2m+1)\pi -\frac{\pi }{4}⟹x=2m\pi +\frac{\pi }{2}$. In $[0,2\pi )$, this is $x=\frac{\pi }{2}$. P: The distinct solutions in $[0,2\pi )$ are $0,\frac{\pi }{4},\frac{\pi }{2},\frac{5\pi }{4}$. Thus, there are 4 solutions. (P)-(4). Q: The solutions in$(0,\pi )$ are $\frac{\pi }{4},\frac{\pi }{2}$. Thus, there are 2 solutions. (Q)-(2). R: The smallest positive solution is$\frac{\pi }{4}$. (R)-(1). S: The sum of all solutions in$[0,2\pi )$ is $0+\frac{\pi }{4}+\frac{\pi }{2}+\frac{5\pi }{4}=\frac{6\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{2}+\frac{\pi }{2}=2\pi$. (S)-(5).