Find all angles, $0^\circ \leq \theta < 360^\circ$ , that satisfy the equation below, to the nearest tenth of a degree. $\newline$ $5\tan^2(\theta)-16 \tan(\theta)=-9\tan(\theta)-2$

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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

5\tan^2(\theta)-16 \tan(\theta)=-9\tan(\theta)-2

Simplify Equation: First, we need to simplify the given equation by moving all terms to one side to set the equation to zero. This will allow us to factor or use other methods to solve for $\tan(\theta)$ . $\newline$ $5\tan^2(\theta) - 16\tan(\theta) + 9\tan(\theta) + 2 = 0$ $\newline$ Simplify the $\tan(\theta)$ terms: $\newline$ $5\tan^2(\theta) - 7\tan(\theta) + 2 = 0$
Factor Quadratic: Next, we factor the quadratic equation in terms of $\tan(\theta)$ . $\$5\tan(\theta) - 2$ (\tan(\theta) - $1$ ) = $0$ \)
Solve for $\tan(\theta)$ : Now, we solve for $\tan(\theta)$ by setting each factor equal to zero. $\newline$ First factor: $5\tan(\theta) - 2 = 0$ $\newline$ $\tan(\theta) = \frac{2}{5}$
Find Angle for $\tan(\frac{2}{5}):$ Second factor: $\tan(\theta) - 1 = 0$ $\newline$ $\tan(\theta) = 1$
Find Angle for $\tan(1)$ : We now find the angles $\theta$ that correspond to the $\tan(\theta)$ values we found. We'll start with $\tan(\theta) = \frac{2}{5}$ . Using a calculator, we find the angle whose tangent is $\frac{2}{5}$ . This angle is in the first quadrant. $\theta \approx \arctan\left(\frac{2}{5}\right) \approx 21.8$ degrees Since tangent is positive in both the first and third quadrants, we also find the angle in the third quadrant by adding $180$ degrees. $\theta \approx 21.8$ degrees $+ 180$ degrees $\approx 201.8$ degrees
Final Angles: Next, we find the angles for $\tan(\theta) = 1$ . The angle whose tangent is $1$ in the first quadrant is $45$ degrees. Since tangent is also positive in the third quadrant, we add $180$ degrees to find the angle in the third quadrant. $\theta = 45$ degrees $+ 180$ degrees $= 225$ degrees
Final Angles: Next, we find the angles for $\tan(\theta) = 1$ . The angle whose tangent is $1$ in the first quadrant is $45$ degrees. Since tangent is also positive in the third quadrant, we add $180$ degrees to find the angle in the third quadrant. $\theta = 45$ degrees $+ 180$ degrees $= 225$ degrees We now have all the angles that satisfy the original equation within the range of $0$ to $360$ degrees. The angles are approximately $21.8$ degrees, $1$ $0$ degrees, and $1$ $1$ degrees.