Answer:
[tex]x=\frac{\pi}{4}[/tex] and [tex]x=-\frac{3\pi}{4}[/tex]
Step-by-step explanation:
We are given that sin x+cos x=cos 2x
We have to solve the given equation for [tex]0\leq x\leq 2\pi[/tex]
[tex] sin x+cos x=cos^2x-sin^2x[/tex]
Because [tex]cos 2x=cos^2-sin^2[/tex]
[tex] sinx+cos x=(sinx +cos x)(sinx-cos x)[/tex]
[tex] 1 =sin x-cos x[/tex]
[tex] sin x=cos x[/tex]
[tex]\frac{sinx }{cos x}=1[/tex]
[tex]tan x=1[/tex]
[tex]tan x=\frac{sinx}{cos x}[/tex]
[tex]tan x=tan\frac{\pi}{4}[/tex]
[tex]x=\frac{\pi}{4}[/tex]
Tan x is positive in I and III quadrant
In III quadrant angle[tex]\theta[/tex] replace by [tex]\theta -\pi[/tex]
Therefore, [tex]tan x=tan (\frac{\pi}{4}-\pi)=tan\frac{\pi-4\pi}{4}=tan\frac{-3\pi}{4}[/tex]
[tex]x=-\frac{3\pi}{4}[/tex]
