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Find all angles, 0^\circ \leq \theta < 360^\circ, that satisfy the equation below, to the nearest tenth of a degree.\newline5tan⁡2(θ)−16tan⁡(θ)=−9tan⁡(θ)−25\tan^2(\theta)-16 \tan(\theta)=-9\tan(\theta)-2

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Q. Find all angles, 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ, that satisfy the equation below, to the nearest tenth of a degree.\newline5tan⁡2(θ)−16tan⁡(θ)=−9tan⁡(θ)−25\tan^2(\theta)-16 \tan(\theta)=-9\tan(\theta)-2
  1. 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⁡(θ)\tan(\theta). \newline5tan⁡2(θ)−16tan⁡(θ)+9tan⁡(θ)+2=05\tan^2(\theta) - 16\tan(\theta) + 9\tan(\theta) + 2 = 0\newlineSimplify the tan⁡(θ)\tan(\theta) terms:\newline5tan⁡2(θ)−7tan⁡(θ)+2=05\tan^2(\theta) - 7\tan(\theta) + 2 = 0
  2. Factor Quadratic: Next, we factor the quadratic equation in terms of tan⁡(θ)\tan(\theta).$5tan⁡(θ)−2\$5\tan(\theta) - 2(\tan(\theta) - 11) = 00\)
  3. Solve for tan⁡(θ)\tan(\theta): Now, we solve for tan⁡(θ)\tan(\theta) by setting each factor equal to zero.\newlineFirst factor: 5tan⁡(θ)−2=05\tan(\theta) - 2 = 0\newlinetan⁡(θ)=25\tan(\theta) = \frac{2}{5}
  4. Find Angle for tan⁡(25):\tan(\frac{2}{5}): Second factor: tan⁡(θ)−1=0\tan(\theta) - 1 = 0\newlinetan⁡(θ)=1\tan(\theta) = 1
  5. Find Angle for tan⁡(1)\tan(1): We now find the angles θ\theta that correspond to the tan⁡(θ)\tan(\theta) values we found. We'll start with tan⁡(θ)=25\tan(\theta) = \frac{2}{5}. Using a calculator, we find the angle whose tangent is 25\frac{2}{5}. This angle is in the first quadrant. θ≈arctan⁡(25)≈21.8\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 180180 degrees. θ≈21.8\theta \approx 21.8 degrees +180+ 180 degrees ≈201.8\approx 201.8 degrees
  6. Final Angles: Next, we find the angles for tan⁡(θ)=1\tan(\theta) = 1. The angle whose tangent is 11 in the first quadrant is 4545 degrees. Since tangent is also positive in the third quadrant, we add 180180 degrees to find the angle in the third quadrant. θ=45\theta = 45 degrees +180+ 180 degrees =225= 225 degrees
  7. Final Angles: Next, we find the angles for tan⁡(θ)=1\tan(\theta) = 1. The angle whose tangent is 11 in the first quadrant is 4545 degrees. Since tangent is also positive in the third quadrant, we add 180180 degrees to find the angle in the third quadrant. θ=45\theta = 45 degrees +180+ 180 degrees =225= 225 degrees We now have all the angles that satisfy the original equation within the range of 00 to 360360 degrees. The angles are approximately 21.821.8 degrees, 1100 degrees, and 1111 degrees.

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