Find the exact solutions of the given equation in the interval [0, 2$\pi$).
$2 \sin ^ { 2 } x + 3 \sin x = - 1$
A) $x = \pi , \frac { 7 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 6 }$
B) $x = \frac { 7 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 4 }$
C) $x = \frac { 7 \pi } { 6 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 6 }$
D) $x = \frac { \pi } { 4 } , \frac { 7 \pi } { 6 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 2 }$
E) $x = 0 , \frac { 7 \pi } { 2 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 2 }$

Question

Find the exact solutions of the given equation in the interval [0, 2$\pi$).  
  $2 \sin ^ { 2 } x + 3 \sin x = - 1$  
A)  $x = \pi , \frac { 7 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 6 }$
B)  $x = \frac { 7 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 4 }$
C)  $x = \frac { 7 \pi } { 6 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 6 }$
D)  $x = \frac { \pi } { 4 } , \frac { 7 \pi } { 6 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 2 }$
E)  $x = 0 , \frac { 7 \pi } { 2 } , \frac { 3 \pi } { 2 } , \frac { 11 \pi } { 2 }$

Quizplus · Accepted Answer

First, rewrite the equation as a quadratic equation in terms of $\sin x$: $2\sin^2x + 3\sin x + 1 = 0$. Factoring or using the quadratic formula, we find $\sin x = -1$ or $\sin x = -\frac{1}{2}$. In the interval $[0, 2\pi)$, $\sin x = -1$ at $x = \frac{3\pi}{2}$, and $\sin x = -\frac{1}{2}$ at $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.

Find the Exact Solutions of the Given Equation in the Interval