Question

Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Give answers in exact form.

Answer 1

Answer:

(a) $x=\frac{5\pi }{3}$

(b)  $x=\frac{\pi }{8}$

Step-by-step explanation:

It is given that $0\leq x\leq 2\pi$

(a) We have given equation $2sinx+\sqrt{3}=0$

$sinx=\frac{-\sqrt{3}}{2}$

$x=sin^{-1}(-0.866)$

$x=\frac{-\pi }{3}=2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$

(b) $tan(2x)=1$

$2x=tan^{-1}1$

$2x=\frac{\pi }{4}$

$x=\frac{\pi }{8}$