Find all angles, $0^{\circ} \leq \theta<360^{\circ}$ , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $-6 \sin ^{2} \theta-9=7 \sin \theta-8$ $\newline$ Answer: $\theta=$

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Algebra 2

Csc, sec, and cot of special angles

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Q. Find all angles,

0^{\circ} \leq \theta<360^{\circ}

, that satisfy the equation below, to the nearest tenth of a degree.

\newline

-6 \sin ^{2} \theta-9=7 \sin \theta-8

\newline

Answer:

\theta=

Rewrite Equation: Rewrite the given equation in standard quadratic form. $\newline$ The given equation is $-6\sin^{2}\theta - 9 = 7\sin \theta - 8$ . To rewrite it in standard quadratic form, we need to move all terms to one side of the equation. $\newline$ $-6\sin^{2}\theta - 7\sin \theta - 9 + 8 = 0$ $\newline$ $-6\sin^{2}\theta - 7\sin \theta - 1 = 0$
Solve Quadratic Equation: Solve the quadratic equation for $\sin \theta$ . We can use the quadratic formula to solve for $\sin \theta$ . The quadratic formula is given by $\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -6$ , $b = -7$ , and $c = -1$ . $\sin \theta = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(-6)(-1)}}{2(-6)}$ $\sin \theta = \frac{7 \pm \sqrt{49 - 24}}{-12}$ $\sin \theta = \frac{7 \pm \sqrt{25}}{-12}$ $\sin \theta = \frac{7 \pm 5}{-12}$
Find Possible Values: Find the two possible values for $\sin \theta$ . We have two possible solutions for $\sin \theta$ : $\sin \theta = \frac{7 + 5}{-12} = \frac{12}{-12} = -1$ $\sin \theta = \frac{7 - 5}{-12} = \frac{2}{-12} = -\frac{1}{6}$
Find Angle for $\sin=-1$ : Find the angles that correspond to $\sin \theta = -1$ . The angle for which $\sin \theta = -1$ is $270°$ .
Find Angle for $\sin=-\frac{1}{6}$ : Find the angles that correspond to $\sin \theta = -\frac{1}{6}$ . To find the angles corresponding to $\sin \theta = -\frac{1}{6}$ , we use the inverse sine function and consider the symmetry of the sine function in the unit circle. $\theta = \arcsin(-\frac{1}{6})$ Since the sine function is negative in the third and fourth quadrants, we need to find the reference angle in the first quadrant and then find the corresponding angles in the third and fourth quadrants. Reference angle = $\arcsin(\frac{1}{6}) \approx 9.6^\circ$ Third quadrant angle = $180^\circ + 9.6^\circ = 189.6^\circ$ Fourth quadrant angle = $360^\circ - 9.6^\circ = 350.4^\circ$