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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 2tan^(2)theta+3tan theta-4=-5 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline2tan⁡2θ+3tan⁡θ−4=−5 2 \tan ^{2} \theta+3 \tan \theta-4=-5 \newlineAnswer: θ= \theta=

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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.\newline2tan⁡2θ+3tan⁡θ−4=−5 2 \tan ^{2} \theta+3 \tan \theta-4=-5 \newlineAnswer: θ= \theta=
  1. Simplify Equation: First, we need to simplify the given equation by moving all terms to one side to set the equation to zero.\newline2tan⁡2θ+3tan⁡θ−4+5=02\tan^{2}\theta + 3\tan \theta - 4 + 5 = 0\newline2tan⁡2θ+3tan⁡θ+1=02\tan^{2}\theta + 3\tan \theta + 1 = 0
  2. Factor Quadratic: Next, we will factor the quadratic equation in terms of tan⁡(θ)\tan(\theta).$2tan⁡(θ)+1\$2\tan(\theta) + 1(\tan(\theta) + 11) = 00\)
  3. Solve for tan⁡(θ)\tan(\theta): Now, we will solve for tan⁡(θ)\tan(\theta) by setting each factor equal to zero.\newlineFirst factor: 2tan⁡θ+1=02\tan \theta + 1 = 0\newlinetan⁡θ=−12\tan \theta = -\frac{1}{2}
  4. Find Angles: Second factor: tan⁡θ+1=0\tan \theta + 1 = 0\newlinetan⁡θ=−1\tan \theta = -1
  5. Use Inverse Tangent: We will find the angles for tan⁡θ=−12\tan \theta = -\frac{1}{2} using the inverse tangent function and considering the periodicity and symmetry of the tangent function.θ=arctan⁡(−12)\theta = \arctan(-\frac{1}{2})This will give us two angles, one in the fourth quadrant and one in the second quadrant.
  6. Calculate Angles: For tan⁡θ=−1\tan \theta = -1, we know that the tangent function is negative in the second and fourth quadrants.\newlineθ=arctan⁡(−1)\theta = \arctan(-1)\newlineThis will give us two angles, one in the fourth quadrant and one in the second quadrant.
  7. Final Angles: We will use a calculator to find the angles to the nearest tenth of a degree.\newlineFor tan⁡θ=−12\tan \theta = -\frac{1}{2}:\newlineθ≈360−arctan⁡(12)≈360−26.6≈333.4\theta \approx 360 - \arctan(\frac{1}{2}) \approx 360 - 26.6 \approx 333.4 degrees (fourth quadrant)\newlineθ≈180−arctan⁡(12)≈180−26.6≈153.4\theta \approx 180 - \arctan(\frac{1}{2}) \approx 180 - 26.6 \approx 153.4 degrees (second quadrant)
  8. Final Angles: We will use a calculator to find the angles to the nearest tenth of a degree.\newlineFor tan⁡θ=−12\tan \theta = -\frac{1}{2}:\newlineθ≈360−arctan⁡(12)≈360−26.6≈333.4\theta \approx 360 - \arctan(\frac{1}{2}) \approx 360 - 26.6 \approx 333.4 degrees (fourth quadrant)\newlineθ≈180−arctan⁡(12)≈180−26.6≈153.4\theta \approx 180 - \arctan(\frac{1}{2}) \approx 180 - 26.6 \approx 153.4 degrees (second quadrant)For tan⁡θ=−1\tan \theta = -1:\newlineθ≈360−arctan⁡(1)≈360−45≈315\theta \approx 360 - \arctan(1) \approx 360 - 45 \approx 315 degrees (fourth quadrant)\newlineθ≈180−arctan⁡(1)≈180−45≈135\theta \approx 180 - \arctan(1) \approx 180 - 45 \approx 135 degrees (second quadrant)
  9. Final Angles: We will use a calculator to find the angles to the nearest tenth of a degree.\newlineFor tan⁡θ=−12\tan \theta = -\frac{1}{2}:\newlineθ≈360−arctan⁡(12)≈360−26.6≈333.4\theta \approx 360 - \arctan(\frac{1}{2}) \approx 360 - 26.6 \approx 333.4 degrees (fourth quadrant)\newlineθ≈180−arctan⁡(12)≈180−26.6≈153.4\theta \approx 180 - \arctan(\frac{1}{2}) \approx 180 - 26.6 \approx 153.4 degrees (second quadrant)For tan⁡θ=−1\tan \theta = -1:\newlineθ≈360−arctan⁡(1)≈360−45≈315\theta \approx 360 - \arctan(1) \approx 360 - 45 \approx 315 degrees (fourth quadrant)\newlineθ≈180−arctan⁡(1)≈180−45≈135\theta \approx 180 - \arctan(1) \approx 180 - 45 \approx 135 degrees (second quadrant)We have found all the angles that satisfy the equation:\newlineθ≈333.4\theta \approx 333.4 degrees, 153.4153.4 degrees, 315315 degrees, and 135135 degrees.

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