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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -5tan^(2)theta-3tan theta+6=-5tan theta+3 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−5tan⁡2θ−3tan⁡θ+6=−5tan⁡θ+3 -5 \tan ^{2} \theta-3 \tan \theta+6=-5 \tan \theta+3 \newlineAnswer: θ= \theta=

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Csc, sec, and cot of special anglesright chevron icon

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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.\newline−5tan⁡2θ−3tan⁡θ+6=−5tan⁡θ+3 -5 \tan ^{2} \theta-3 \tan \theta+6=-5 \tan \theta+3 \newlineAnswer: θ= \theta=
  1. Simplify Equation: First, let's simplify the given equation by moving all terms to one side to set the equation to zero.\newline−5tan⁡2θ−3tan⁡θ+6+5tan⁡θ−3=0-5\tan^{2}\theta - 3\tan \theta + 6 + 5\tan \theta - 3 = 0\newlineCombine like terms:\newline−5tan⁡2θ+2tan⁡θ+3=0-5\tan^{2}\theta + 2\tan \theta + 3 = 0
  2. Factor Quadratic Equation: Next, we will factor the quadratic equation in terms of tan⁡θ\tan \theta. We are looking for two numbers that multiply to −5×3=−15-5\times3 = -15 and add up to 22. These numbers are 55 and −3-3. So we can write the equation as: (−5tan⁡θ+3)(tan⁡θ+1)=0(-5\tan \theta + 3)(\tan \theta + 1) = 0
  3. Solve for tan⁡θ\tan \theta: Now, we will solve each factor for tan⁡θ\tan \theta:\newlineFirst factor: −5tan⁡θ+3=0-5\tan \theta + 3 = 0\newlinetan⁡θ=35\tan \theta = \frac{3}{5}
  4. Find Angles for tan⁡θ=35\tan \theta = \frac{3}{5}: Second factor: tan⁡θ+1=0\tan \theta + 1 = 0\newlinetan⁡θ=−1\tan \theta = -1
  5. Find Angles for tan⁡θ=−1\tan \theta = -1: We will now find the angles for tan⁡θ=35\tan \theta = \frac{3}{5}. The arctangent of 35\frac{3}{5} gives us the reference angle. We use a calculator to find this angle: θ=arctan⁡(35)\theta = \arctan(\frac{3}{5}) θ≈30.96\theta \approx 30.96 degrees Since the tangent function is positive in the first and third quadrants, we find the angles in those quadrants: For the first quadrant: θ≈30.96\theta \approx 30.96 degrees For the third quadrant: θ≈180+30.96≈210.96\theta \approx 180 + 30.96 \approx 210.96 degrees
  6. Find Angles for tan⁡θ=−1\tan \theta = -1: We will now find the angles for tan⁡θ=35\tan \theta = \frac{3}{5}. The arctangent of 35\frac{3}{5} gives us the reference angle. We use a calculator to find this angle: θ=arctan⁡(35)\theta = \arctan(\frac{3}{5}) θ≈30.96\theta \approx 30.96 degrees Since the tangent function is positive in the first and third quadrants, we find the angles in those quadrants: For the first quadrant: θ≈30.96\theta \approx 30.96 degrees For the third quadrant: θ≈180+30.96≈210.96\theta \approx 180 + 30.96 \approx 210.96 degreesNext, we find the angles for tan⁡θ=−1\tan \theta = -1. The arctangent of −1-1 gives us the reference angle. We use a calculator to find this angle: θ=arctan⁡(−1)\theta = \arctan(-1) tan⁡θ=35\tan \theta = \frac{3}{5}00 degrees Since the tangent function is negative in the second and fourth quadrants, we find the angles in those quadrants: For the second quadrant: tan⁡θ=35\tan \theta = \frac{3}{5}11 degrees For the fourth quadrant: tan⁡θ=35\tan \theta = \frac{3}{5}22 degrees

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