need help solving
$\displaystyle \displaystyle \cos 2\theta =\frac{\sqrt{3}}{2}$
Make $\displaystyle \displaystyle 2\theta =x$ therefore
$\displaystyle \displaystyle \cos x =\frac{\sqrt{3}}{2}$
Using the 30-60-90 special triangle
$\displaystyle \displaystyle x =\frac{\pi}{6}\implies 2\theta = \frac{\pi}{6} \implies \theta = \frac{\pi}{12}$
Now what other quadrants is $\displaystyle \cos $ positive in? and how many solutions are we looking for?
If you want to use a method that does not require as much analysis...
$\displaystyle \displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}$
$\displaystyle \displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n$ where $\displaystyle \displaystyle n$ is an integer
$\displaystyle \displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n$
$\displaystyle \displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n$.
So the solutions in the domain $\displaystyle \displaystyle [0, 360]$ are
$\displaystyle \displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}$.
If you want to use a method that does not require as much analysis...
$\displaystyle \displaystyle \cos{2\theta} = \frac{\sqrt{3}}{2}$
$\displaystyle \displaystyle 2\theta = \{30^{\circ}, 360^{\circ} - 30^{\circ}\} + 360^{\circ}n$ where $\displaystyle \displaystyle n$ is an integer
$\displaystyle \displaystyle 2\theta = \{30^{\circ}, 330^{\circ}\} + 360^{\circ}n$
$\displaystyle \displaystyle \theta = \{15^{\circ}, 165^{\circ}\} + 180^{\circ}n$.
So the solutions in the domain $\displaystyle \displaystyle [0, 360]$ are
$\displaystyle \displaystyle \{15^{\circ}, 165^{\circ}, 195^{\circ}, 345^{\circ}\}$.