Answer:
[tex]x=0,~x=\frac{2\pi }{3},\:x=\frac{4\pi }{3},~x=2\pi[/tex]
Step-by-step explanation:
The given equation:
[tex]\cos(x)=2\cos^2(x)-1[/tex]
Let [tex]\cos(x)[/tex] be [tex]y[/tex]:
[tex]y=2y^2-1[/tex]
Rewrite as:
[tex]2y^2-y-1=0[/tex]
After solving quadratic equation, the solutions are:
[tex]y=1~~~~y=-\frac{1}{2}[/tex]
Substitute back [tex]\cos(x)[/tex], it follows:
[tex]\cos(x)=1~~~~\cos(x)=-\frac12[/tex]
For the first equation, the solution is [tex]x=0[/tex] and [tex]x=2\pi[/tex].
For the second equation, general solution is:
[tex]x=\frac{2\pi }{3}+2\pi n,\:x=\frac{4\pi }{3}+2\pi n[/tex]
But for the given interval, we must substitute n=0 into the equations so that the values of x must be within interval. Therefore:
[tex]x=\frac{2\pi }{3},\:x=\frac{4\pi }{3}[/tex]
So, the answers are:
[tex]x=0,~x=\frac{2\pi }{3},\:x=\frac{4\pi }{3},~x=2\pi[/tex]
