Use Identities to Solve Trigonometric Equations
Solve the equation on the interval [0, 2π).
-$2 \cos ^ { 2 } x + \sin x - 2 = 0$
A) $0 , \pi , \frac { \pi } { 6 } , \frac { 5 \pi } { 6 }$
B) $\frac { \pi } { 2 } , \frac { 3 \pi } { 2 } , \frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$
C) $\frac { \pi } { 6 } , \frac { 5 \pi } { 6 }$
D) $\frac { \pi } { 3 } , \frac { 2 \pi } { 3 }$

Question

Quizplus · Accepted Answer

Using the identity $ \cos^2x = 1 - \sin^2x $, the equation can be rewritten as $2(1 - \sin^2x) + \sin x - 2 = 0$, simplifying to $2 - 2\sin^2x + \sin x - 2 = 0$, or $2\sin^2x - \sin x = 0$. Factoring gives $\sin x(2\sin x - 1) = 0$, leading to $\sin x = 0$ and $\sin x = \frac{1}{2}$. The solutions within the interval [0, 2π) for $\sin x = 0$ are $x = 0$ and $x = \pi$, and for $\sin x = \frac{1}{2}$, the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.

Use Identities to Solve Trigonometric Equations Solve the Equation on the Interval

Use Identities to Solve Trigonometric Equations
Solve the Equation on the Interval