Find the exact solutions of the given equation in the interval [0,2π). $$2 \sin ^ { 2 } x + 3 \sin x = - 1$$
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 [0,2π). ​​ $$2 \sin ^ { 2 } x + 3 \sin x = - 1$$ ​
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 given equation as a quadratic equation in terms of $\sin x$: $2\sin^2x + 3\sin x + 1 = 0$. Factoring this equation, we get $(2\sin x + 1)(\sin x + 1) = 0$, which gives us $\sin x = -\frac{1}{2}$ and $\sin x = -1$. Solving for $x$ in the interval $[0, 2\pi)$, we find $x = \frac{7\pi}{6}, \frac{11\pi}{6}$ (for $\sin x = -\frac{1}{2}$) and $x = \frac{3\pi}{2}$ (for $\sin x = -1$).

Find the Exact Solutions of the Given Equation in the Interval