# Appendix D - Trigonometry - D Exercises - Page A33: 74

$0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$

#### Work Step by Step

$2cosx+1>0$ Here, $2cosx+1=0$ gives $cosx=-\frac{1}{2}$ Since, $cosx<0$ on second and third quadrant thus, x must lie on $\frac{2\pi}{3},\frac{4 \pi}{3}$ Hence, the solution for the given equation is $0\leq x\leq\frac{2\pi}{3},\frac{4 \pi}{3}\leq x\leq 2\pi$

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