Question

Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Round to three decimal places when necessary.

Answer 1

Answer:

(A) $\frac{\pi }{3}$

(b) $1.768\pi$

Step-by-step explanation:

It is given in the question that  $0\leq x\leq 2\pi$

(a) We have given equation $2cosx-1=0$

$cosx=\frac{1}{2}$

$x=cos^{-1}\frac{1}{2}$

$x=\frac{\pi }{3}$

(b) $3sinx+2=0$

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

$x=sin^{-1}(-0.666)$

$x=-41.75^{\circ}=360-41.75=318.25^{\circ}=1.768\pi$