Answer:
[tex]x=0,\pi,2\pi[/tex]
Explanation:
[tex]cos^{2}(x)-cos(2x)=0[/tex]
Firstly, we will simplify the double angle formulae
[tex]cos(2x)=cos(x+x)\\cos(x+x)=cos(x)cos(x)-sin(x)sin(x)\\cos(x+x)=cos^{2}(x)-sin^{2}(x)[/tex]
Now, we can substitute our value for cos(2x) into our original equation.
[tex]cos^{2}(x)-(cos^{2}(x)-sin^{2}(x))=0\\cos^2(x)-cos^2(x)+sin^2(x)=0\\sin^2(x)=0\\sin(x)=0\\x=arcsin(0)\\x=0,\pi,2\pi[/tex]
(Solutions in the range [tex]0 \leq x \leq 2\pi[/tex])
