Solve the equation in the interval [0°, 360°). Give solutions to the nearest tenth, if necessary.
-$\sin ^ { 2 } \theta - \sin \theta - 12 = 0$
A) $\left\{ 45 ^ { \circ } , 135 ^ { \circ } \right\}$
B) $\left\{ 45 ^ { \circ } , 315 ^ { \circ } \right\}$
C) $\varnothing$
D) $\left\{ 45 ^ { \circ } \right\}$

Question

Quizplus · Accepted Answer

This equation can be solved by treating it as a quadratic equation in terms of $\sin \theta$. Let $x = \sin \theta$, then the equation becomes $x^2 - x - 12 = 0$. Factoring this gives $(x-4)(x+3) = 0$, so $x = 4$ or $x = -3$. Since the sine function can only take values in the range $[-1, 1]$, there are no real solutions for $\theta$ in the interval [0°, 360°) that satisfy the original equation.

Solve the Equation in the Interval [0°, 360°) sin⁡2θ−sin⁡θ−12=0\sin ^ { 2 } \theta - \sin \theta - 12 = 0sin2θ−sinθ−12=0

Solve the Equation in the Interval [0°, 360°) $\sin ^ { 2 } \theta - \sin \theta - 12 = 0$