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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. -6tan^(2)theta-16 tan theta=-9tan theta+2 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline−6tan⁡2θ−16tan⁡θ=−9tan⁡θ+2 -6 \tan ^{2} \theta-16 \tan \theta=-9 \tan \theta+2 \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−6tan⁡2θ−16tan⁡θ=−9tan⁡θ+2 -6 \tan ^{2} \theta-16 \tan \theta=-9 \tan \theta+2 \newlineAnswer: θ= \theta=
  1. Rewrite Equation: First, we need to rewrite the given equation in a standard quadratic form by moving all terms to one side of the equation.\newline−6tan⁡2θ−16tan⁡θ+9tan⁡θ−2=0-6\tan^{2}\theta - 16 \tan \theta + 9\tan \theta - 2 = 0\newlineThis simplifies to:\newline−6tan⁡2θ−7tan⁡θ−2=0-6\tan^{2}\theta - 7 \tan \theta - 2 = 0
  2. Factor Quadratic: Next, we attempt to factor the quadratic equation, if possible. If not, we will use the quadratic formula to find the solutions for tan⁡θ\tan \theta. Factoring the quadratic equation gives us: (−2tan⁡θ−1)(3tan⁡θ+2)=0(-2\tan \theta - 1)(3\tan \theta + 2) = 0
  3. Solve for tan⁡θ\tan \theta: Now, we set each factor equal to zero and solve for tan⁡θ\tan \theta. First factor: −2tan⁡θ−1=0-2\tan \theta - 1 = 0 tan⁡θ=−12\tan \theta = -\frac{1}{2} Second factor: 3tan⁡θ+2=03\tan \theta + 2 = 0 tan⁡θ=−23\tan \theta = -\frac{2}{3}
  4. Find Angles: We need to find the angles θ\theta that correspond to tan⁡θ=−12\tan \theta = -\frac{1}{2} and tan⁡θ=−23\tan \theta = -\frac{2}{3} within the interval 0^\circ \leq \theta < 360^\circ. For tan⁡θ=−12\tan \theta = -\frac{1}{2}, the angles are in the second and fourth quadrants. We find the reference angle by taking the arctan⁡\arctan of 12\frac{1}{2} and then subtract from 180∘180^\circ for the second quadrant and add to 360∘360^\circ for the fourth quadrant. Reference angle = arctan⁡(12)≈26.6∘\arctan(\frac{1}{2}) \approx 26.6^\circ Second quadrant angle = tan⁡θ=−12\tan \theta = -\frac{1}{2}00 Fourth quadrant angle = tan⁡θ=−12\tan \theta = -\frac{1}{2}11
  5. Reference Angles: For tan⁡θ=−23\tan \theta = -\frac{2}{3}, the angles are also in the second and fourth quadrants. We find the reference angle by taking the arctan⁡(23)\arctan \left(\frac{2}{3}\right) and then subtract from 180∘180^\circ for the second quadrant and add to 360∘360^\circ for the fourth quadrant.\newlineReference angle = arctan⁡(23)≈33.7∘\arctan\left(\frac{2}{3}\right) \approx 33.7^\circ\newlineSecond quadrant angle = 180∘−33.7∘≈146.3∘180^\circ - 33.7^\circ \approx 146.3^\circ\newlineFourth quadrant angle = 360∘−33.7∘≈326.3∘360^\circ - 33.7^\circ \approx 326.3^\circ
  6. Final Angles: We now have four angles that satisfy the original equation: 153.4∘153.4^\circ, 333.4∘333.4^\circ, 146.3∘146.3^\circ, and 326.3∘326.3^\circ. These are the angles to the nearest tenth of a degree.

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