Find all solutions to the equation.
-$4 \sin ^ { 2 } x - 4 \sin x + 1 = 0$
A) $\left\{ \frac { \pi } { 6 } + 2 \mathrm { n } \pi , \frac { 7 \pi } { 6 } + 2 \mathrm { n } \pi \mid \mathrm { n } = 0 , \pm 1 , \pm 2 , \ldots \right\}$
B) $\left\{ \frac { \pi } { 6 } + 2 \mathrm { n } \pi , \frac { 5 \pi } { 6 } + 2 \mathrm { n } \pi \mid \mathrm { n } = 0 , \pm 1 , \pm 2 , \ldots \right\}$
C) $\left\{ \frac { \pi } { 3 } + 2 \mathrm { n } \pi , \frac { 5 \pi } { 3 } + 2 \mathrm { n } \pi \mid \mathrm { n } = 0 , \pm 1 , \pm 2 , \ldots . \right\}$
D) $\left\{ \frac { \pi } { 6 } + 2 \mathrm { n } \pi , \frac { 11 \pi } { 6 } + 2 \mathrm { n } \pi \mid \mathrm { n } = 0 , \pm 1 , \pm 2 , \ldots . \right\}$

Question

Quizplus · Accepted Answer

The given equation is a quadratic in $\sin x$: $4\sin^2x - 4\sin x + 1 = 0$. Factoring or using the quadratic formula, we find $\sin x = \frac{1}{2}$. The solutions to $\sin x = \frac{1}{2}$ are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$, plus any integer multiple of $2\pi$ to account for the periodic nature of the sine function.

Find All Solutions to the Equation 4sin⁡2x−4sin⁡x+1=04 \sin ^ { 2 } x - 4 \sin x + 1 = 04sin2x−4sinx+1=0

Find All Solutions to the Equation $4 \sin ^ { 2 } x - 4 \sin x + 1 = 0$