Answer
$\theta=60^{\circ}+180^{\circ}k,120^{\circ}+180^{\circ}k,180^{\circ}k$,
where k is any integer.
Work Step by Step
$2(\cos2\theta)^{2}-\cos2\theta-1=0$
or, $(2cos2\theta+1)(\cos2\theta-1)=0$
either $cos2\theta=-\frac{1}{2}$ or $1$
For, $\cos2\theta=-\frac{1}{2}$,
$2\theta=\cos^{-1}(-\frac{1}{2})$,
Since, $\cos2\theta$ is negative in both quadrants 2 and 3,
$2\theta=180^{\circ}-60^{\circ}=120^{\circ} \,\,or \,\,180^{\circ}+60^{\circ}=240^{\circ},$
To find all values of $2\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of cos function is $360^{\circ}$.
So, $2\theta=120^{\circ}+360^{\circ}k\,\,or \,\,240^{\circ}+360^{\circ}k,$
So, $\theta=60^{\circ}+180^{\circ}k\,\,or \,\,120^{\circ}+180^{\circ}k,$
For, $\cos2\theta=1$,
$2\theta=\cos^{-1}(1)$,
$2\theta=0^{\circ}$
To find all values of $2\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of cos function is $360^{\circ}$.
So, $2\theta=0^{\circ}+360^{\circ}k\,\,$
So, $\theta=180^{\circ}k\,\,$
So, all degree solutions are
$\theta=60^{\circ}+180^{\circ}k,120^{\circ}+180^{\circ}k,180^{\circ}k$
