Check my solution to this trig inequality

Question

Problem $1.88$ :

Solve $$\cos x\lt \frac{\sqrt{3}}{2},\qquad x \in [0,2\pi]$$ I found the set of solutions to be $S=[0,2\pi]-\left[\dfrac{\pi}{6},\dfrac{11\pi}{6}\right]$ Is this correct? Thank you.

Answer 1

Actually I found $$\frac{\pi}{6} \lt x \lt \frac{11\pi}{6}$$ so $$x \in \left]\frac{\pi}{6},\frac{11\pi}{6}\right[$$ The set of solution you found $$S=[0,2\pi]-\left[\dfrac{\pi}{6},\dfrac{11\pi}{6}\right]$$ is the answer to the inequality $$\cos x\gt \frac{\sqrt{3}}{2},\qquad x \in [0,2\pi]$$