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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=13 tan theta+8 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline6tan⁡2θ=13tan⁡θ+8 6 \tan ^{2} \theta=13 \tan \theta+8 \newlineAnswer: θ= \theta=

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Csc, sec, and cot of special anglesright chevron icon

Full solution

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.\newline6tan⁡2θ=13tan⁡θ+8 6 \tan ^{2} \theta=13 \tan \theta+8 \newlineAnswer: θ= \theta=
  1. Write Standard Quadratic Form: Write the given equation in the standard quadratic form by moving all terms to one side.\newline6tan⁡2(θ)−13tan⁡(θ)−8=06\tan^2(\theta) - 13\tan(\theta) - 8 = 0
  2. Factor Quadratic Equation: Factor the quadratic equation to find the values of tan⁡(θ)\tan(\theta).(2tan⁡(θ)+1)(3tan⁡(θ)−8)=0(2\tan(\theta) + 1)(3\tan(\theta) - 8) = 0
  3. Set and Solve for tan⁡(θ)\tan(\theta): Set each factor equal to zero and solve for tan⁡(θ)\tan(\theta).2tan⁡(θ)+1=02\tan(\theta) + 1 = 0 or 3tan⁡(θ)−8=03\tan(\theta) - 8 = 0
  4. Solve for tan⁡(θ)=−12\tan(\theta) = -\frac{1}{2}: Solve the first equation for tan⁡(θ)\tan(\theta).\newlinetan⁡(θ)=−12\tan(\theta) = -\frac{1}{2}
  5. Solve for tan⁡(θ)=83\tan(\theta) = \frac{8}{3}: Solve the second equation for tan⁡(θ)\tan(\theta).\newlinetan⁡(θ)=83\tan(\theta) = \frac{8}{3}
  6. Find Angles for tan⁡(θ)=−12\tan(\theta) = -\frac{1}{2}: Find the angles that correspond to tan⁡(θ)=−12\tan(\theta) = -\frac{1}{2} in the interval [0∘,360∘)[0^\circ, 360^\circ). The angles are approximately 333.4∘333.4^\circ and 153.4∘153.4^\circ.
  7. Find Angles for tan⁡(θ)=83\tan(\theta) = \frac{8}{3}: Find the angles that correspond to tan⁡(θ)=83\tan(\theta) = \frac{8}{3} in the interval [0∘,360∘)[0^\circ, 360^\circ).\newlineThe angles are approximately 69.4∘69.4^\circ and 249.4∘249.4^\circ.

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