Simple Trigonometric Equation I am asked to solve the trigonometric equation $2cos \theta = \sqrt 3$
I rearrange it to $cos\theta = \frac{\sqrt3}{2}$
Now, at this point I am not sure what to do? Can someone describe to me the steps required? What identities do I need to know or be familiar with? 
Cheers.
 A: First, there are some exact values of the sine and cosine you should always remember:

Using this table, we identify the first solution to be $\frac{\pi}{6}$.
Now, recall there are two solutions:

How do we find the second solution? 
Using the unit circle, we ask ourselves the following: in which quadrant of the unit circle is the cosine (which is represented by the $x$-value) positive? (since $\frac{\sqrt{3}}{2}$ is positive).
Well, in the first quadrant (the value we already found) and in the fourth quadrant:

The angle in the fourth quadrant is found by: $2\pi-\frac{\pi}{6}=\frac{11\pi}{6}$.
Using periodicity, we find the general answers to be:
$x_1=\frac{\pi}{6}+k\cdot2\pi \\ x_2=\frac{11\pi}{6}+k\cdot2\pi$
A: You can take the $\arccos$ (principal value) of both sides of the equation, yielding
$$\arccos\cos\theta=\theta^*=\arccos\frac{\sqrt3}2=\frac\pi6.$$
But this is not enough as there can be other values of $\theta$ that have the same cosine:


*

*the cosine function is periodic: $\cos\theta=\cos(\theta+2k\pi)$,

*the cosine function is even: $\cos\theta=\cos(-\theta)$.
Put together, this gives
$$\theta=\pm\theta^*+2k\pi.$$



