Answer and Explanation:
Given : Equation \(2\cos^2x = \sin2x\)
To find : Solve the given equation ?
Solution :
Equation \(2\cos^2x = \sin2x\)
\(2\cos^2x-\sin2x=0\)
Open identity, \(\sin 2x=2\sin x\cos x\)
\(2\cos^2x-2\sin x\cos x=0\)
\(2\cos x(\cos x-\sin x)=0\)
\(2\cos x=0,(\cos x-\sin x)=0\)
1) When \(2\cos x=0\)
\(\cos x=0\)
General solution - \(x=\frac{\pi}{2}+n\pi\)
From \((0,2\pi)\) the solution is \((\frac{\pi}{2},0),(\frac{3\pi}{2},0)\)
2) When \((\cos x-\sin x)=0\)
\(\cos x=\sin x\)
\(1=\frac{\sin x}{\cos x}\)
\(\tan x=1\)
General solution - \(x=\frac{\pi}{4}+n\pi\)
From \((0,2\pi)\) the solution is \((\frac{\pi}{4},0),(\frac{5\pi}{4},0)\)
