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Find all angles, 0^(@) <= theta < 360^(@), that satisfy the equation below, to the nearest tenth of a degree. 25csc^(2)theta-9=0 Answer: theta=

Find all angles, 0^{\circ} \leq \theta<360^{\circ} , that satisfy the equation below, to the nearest tenth of a degree.\newline25csc⁡2θ−9=0 25 \csc ^{2} \theta-9=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.\newline25csc⁡2θ−9=0 25 \csc ^{2} \theta-9=0 \newlineAnswer: θ= \theta=
  1. Isolate csc⁡2θ\csc^2\theta term: Start by isolating the csc⁡2θ\csc^2\theta term in the equation 25csc⁡2θ−9=025\csc^2\theta - 9 = 0. Add 99 to both sides of the equation to get: 25csc⁡2θ=925\csc^2\theta = 9
  2. Add 99 to both sides: Divide both sides of the equation by 2525 to solve for csc⁡2θ\csc^2\theta.\newlinecsc⁡2θ=925\csc^2\theta = \frac{9}{25}
  3. Divide by 2525: Take the square root of both sides to solve for csc(θ)csc(\theta). Remember that taking the square root gives us two solutions: one positive and one negative.\newlinecsc(θ)=±925csc(\theta) = \pm\sqrt{\frac{9}{25}}\newlinecsc(θ)=±35csc(\theta) = \pm\frac{3}{5}
  4. Take square root: Since csc⁡(θ)\csc(\theta) is the reciprocal of sin⁡(θ)\sin(\theta), we can write:\newlinesin⁡(θ)=±1(3/5)\sin(\theta) = \pm\frac{1}{(3/5)}\newlinesin⁡(θ)=±53\sin(\theta) = \pm\frac{5}{3}\newlineHowever, the sine function has a range of [−1,1][-1, 1], so sin⁡(θ)\sin(\theta) cannot be ±53\pm\frac{5}{3}. This indicates a mistake has been made.

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