Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $16 \sin ^{2} \theta-9=0$ $\newline$ Answer: $\theta=$

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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

16 \sin ^{2} \theta-9=0

\newline

Answer:

\theta=

Solve for $\sin^2(\theta)$ : Solve the equation for $\sin^2(\theta)$ .
$16\sin^2(\theta) - 9 = 0$
Add $9$ to both sides of the equation.
$16\sin^2(\theta) = 9$
Divide both sides by $16$ .
$\sin^2(\theta) = \frac{9}{16}$
Take the square root of both sides.
$\sin(\theta) = \pm\sqrt{\frac{9}{16}}$
$\sin(\theta) = \pm\frac{3}{4}$
Positive Value of $\sin(\theta)$ : Find the angles for the positive value of $\sin(\theta)$ . $\sin(\theta) = \frac{3}{4}$ Since $\sin$ is positive, $\theta$ could be in the first or second quadrant.Use the inverse sine function to find the angle in the first quadrant. $\theta = \sin^{-1}(\frac{3}{4})$ $\theta \approx 48.6^\circ$
Second Quadrant Angle: Find the angle in the second quadrant. $\newline$ Since the sine function is symmetric with respect to the y-axis, the second quadrant angle will be $180° - \theta$ . $\newline$ $\theta = 180° - 48.6°$ $\newline$ $\theta \approx 131.4°$
Negative Value of $\sin(\theta)$ : Find the angles for the negative value of $\sin(\theta)$ . $\sin(\theta) = -\frac{3}{4}$ Since $\sin$ is negative, $\theta$ could be in the third or fourth quadrant.For the third quadrant, the reference angle is the same as the first quadrant, but we add $180^\circ$ . $\theta = 180^\circ + 48.6^\circ$ $\theta \approx 228.6^\circ$
Fourth Quadrant Angle: Find the angle in the fourth quadrant. $\newline$ For the fourth quadrant, we subtract the reference angle from $360°$ . $\newline$ $\theta = 360° - 48.6°$ $\newline$ $\theta \approx 311.4°$