The number of solutions of x+ x = 2 x in [0,2) is:

Question

The number of solutions of $	an x+\sec x = 2\cos x$ in $[0,2\pi)$ is:

Accepted Answer

$$\tan x+\sec x = 2\cos x \implies \sin x+1 = 2\cos^2x = 2(1-\sin^2x) \implies 2\sin^2x+\sin x-1=0$$

$$(2\sin x-1)(\sin x+1)=0 \implies \sin x = \frac12 \text{ or } -1$$

$$x = 30^\circ,\ 150^\circ,\ 270^\circ$$

3 solutions.

Solutions · 1