Solve $\cos x>1/2$ for $-180Solve 
$$\cos x>\frac12\quad\text{for}\quad-180^{\circ}<x<180^{\circ}\;.$$
Hey guys, Ive got the solution to this question although I cant seem to figure out how the textbook did it. Can someone please explain this question thanks
 A: Recall that, in degrees, we know that $$\cos(\pm 60^\circ) = \dfrac 12$$
It's easy to see, using the unit circle, that when  $$-60^\circ < x < 60^\circ,\quad \cos x \gt \dfrac 12$$
See, e.g., the graphic below (compliments of Wikipedia) of the unit circle, with the values of $(\cos \theta, \sin \theta)$ listed along the circumference of the circle. 
Can you see at what points on the unit circle, the x-coordinates are greater than $\frac 12,$ and for which angles the cosine values are greater than $\frac 12$?:

Note that, e.g., $-60^\circ \sim 300^\circ, -180 = 180^\circ$
A: Hint
Draw a picture and recall that
$$\cos\left(\pm\frac{\pi}{3}\right)=\frac{1}{2}$$
A: Consider a point $P=\langle x,y\rangle$ on the unit circle $x^2+y^2=1$. If $\theta$ is the angle from the positive $x$-axis to the ray from the origin through $P$, then $\cos\theta=x$ and $\sin\theta=y$. (This is one of the three basic ways of looking at the sine and cosine functions.) You want to find the values of $\theta$ satisfying $\cos\theta>\frac12$ and $-180<\theta<180$. Draw a picture. You should know that $\cos(-60°)=\cos 60°=\frac12$. What points on the unit circle have $x$-coordinates bigger than $\frac12$ and therefore correspond to angles $\theta$ whose cosines are bigger than $\frac12$?

A: If you know the graph of the cosine function, then you can see things easily: the question is asking what part of the graph is above the horizontal line $y=1/2$.
