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

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newlinesin⁡2θ−4=0 \sin ^{2} \theta-4=0 \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.\newlinesin⁡2θ−4=0 \sin ^{2} \theta-4=0 \newlineAnswer: θ= \theta=
  1. Find sin⁡2θ\sin^{2}\theta: Solve the equation sin⁡2θ−4=0\sin^{2}\theta - 4 = 0 for sin⁡2θ\sin^{2}\theta.\newlinesin⁡2θ=4\sin^{2}\theta = 4
  2. Take square root: Take the square root of both sides to solve for sin⁡(θ)\sin(\theta).sin⁡(θ)=±4\sin(\theta) = \pm\sqrt{4}sin⁡(θ)=±2\sin(\theta) = \pm2Since the sine of an angle cannot be greater than 11 or less than −1-1, this equation has no solution.

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