Inequality’s solutions length? What does “interval whose length is equal to ...” mean?
 In this question,
All solutions of the inequality,
$$\cos^2\big(\frac{x}{2}\big) ≥ \sin^2\big(\frac{x}{2}\big)-\big(\frac{1}{2}\big) $$
That belongs to, $x\in[-\pi; \pi]$. 
Interval length is equal to ...
.
I don’t understand what’s the question?? If anyone could help I would appreciate it A LOT!! 
 A: Use that $$\cos(x)=\cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)$$
A: I'm not sure what the question is, but I guess that you've been asked

What's the length of the interval, in which the inequality $$\cos^2\big(\frac{x}{2}\big) ≥ \sin^2\big(\frac{x}{2}\big)-\big(\frac{1}{2}\big) $$ holds for $x\in [-\pi; \pi]$

If so, just as @Dr. Sonnhard Graubner pointed out, using the identity $\cos(2\varphi)=\cos^2(\varphi)-\sin^2(\varphi)$: $$\cos (x)=\cos^2\big(\frac{x}{2}\big)-\sin^2\big(\frac{x}{2}\big)$$ And applying this in your inequality $$\cos^2\big(\frac{x}{2}\big) ≥ \sin^2\big(\frac{x}{2}\big)-\big(\frac{1}{2}\big) \iff \cos^2\big(\frac{x}{2}\big)- \sin^2\big(\frac{x}{2}\big)=\cos(x) ≥ -\big(\frac{1}{2}\big) $$
Now you just have to figure out for which values of $x\in [-\pi; \pi]$ the inequality holds

 Observe that $\cos\big(\frac{\pi}{3}\big)=0.5$ and since $\cos(\pi-x)=-\cos(x)$ $$\cos \big(\frac{2\pi}{3}\big)=-\frac{1}{2}$$ Can you continue from here?

A: Hint: Use the fact that $\cos(x)>-1/2$ for $x\in[-{2\pi\over 3},{3\pi\over 3}]$
