Solve the equation in the interval [0°, 360°). Give solutions to the nearest tenth, if necessary.
-$\sin ^ { 2 } \theta - \sin \theta - 12 = 0$

Question

Quizplus · Accepted Answer

The given equation is a quadratic in terms of $\sin \theta$. Solving it as such, we find that the solutions for $\sin \theta$ fall outside the range [-1,1], which means there are no real angles $\theta$ in the interval [0°, 360°) that satisfy the 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°) $\sin ^ { 2 } \theta - \sin \theta - 12 = 0$