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