Find all solutions of the following equation in the interval $[ 0,2 \pi )$.
$2 \cos ^ { 2 } x = 2 + \sin x$
A) $x = 0 , \frac { \pi } { 4 } , \frac { 3 \pi } { 4 }$
B) $x = 0 , \frac { 3 \pi } { 4 } , \frac { 4 \pi } { 3 } , \frac { 11 \pi } { 6 }$
C) $x = \frac { \pi } { 2 } , \frac { 2 \pi } { 3 } , \frac { 4 \pi } { 3 } , \frac { 3 \pi } { 2 }$
D) $x = 0 , \pi , \frac { 7 \pi } { 6 } , \frac { 11 \pi } { 6 }$
E) $x = \frac { \pi } { 2 } , \frac { 3 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 7 \pi } { 6 }$

Question

Quizplus · Accepted Answer

First, rewrite the equation using the identity $ \cos^2x = 1 - \sin^2x $ to get $ 2(1 - \sin^2x) = 2 + \sin x $. Simplify to get a quadratic equation in $ \sin x $, solve for $ \sin x $, and find the corresponding $ x $ values in the given interval.

Find All Solutions of the Following Equation in the Interval [0,2π)[ 0,2 \pi )[0,2π)

Find All Solutions of the Following Equation in the Interval $[ 0,2 \pi )$