Answer:
θ∈{[tex]\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}[/tex]}
Explanation:
The given equation is
[tex]sin(2\theta )-cos(2\theta )=0[/tex]
[tex]\Rightarrow sin(2\theta )=cos(2\theta )\\\\\therefore \frac{sin(2\theta )}{cos(2\theta )}=1\\\\tan(2\theta )=1\\\\\therefore 2\theta =n\pi +\frac{\pi}{4}\\\\\therefore \theta =\frac{n\pi }{2}+\frac{\pi }{8}[/tex]
Applying values on 'n' we obtain values of θ that beling to [0,2π)
For n=0, θ=[tex]\frac{\pi }{8}[/tex]
For n=1, θ =[tex]\frac{5\pi }{8}[/tex]
For n=2,θ =[tex]\frac{9\pi }{8}[/tex]
For n=3,θ =[tex]\frac{13\pi }{8}[/tex]
