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Find all solutions with −π2≤θ≤π2 - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} . Give the exact answer(s) in simplest form. If there are multiple answers, separate them with commas. 9csc⁡(θ)−18=09\csc(\theta)-18=0

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Q. Find all solutions with −π2≤θ≤π2 - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} . Give the exact answer(s) in simplest form. If there are multiple answers, separate them with commas. 9csc⁡(θ)−18=09\csc(\theta)-18=0
  1. Simplify Equation: Step 11: Simplify the equation 9csc⁡(θ)−18=09\csc(\theta) - 18 = 0 to find csc⁡(θ)\csc(\theta).\newlineCalculation: 9csc⁡(θ)−18=09\csc(\theta) - 18 = 0\newline 9csc⁡(θ)=189\csc(\theta) = 18\newline csc⁡(θ)=189\csc(\theta) = \frac{18}{9}\newline csc⁡(θ)=2\csc(\theta) = 2
  2. Convert to Sin: Step 22: Convert csc⁡(θ)\csc(\theta) to sin⁡(θ)\sin(\theta) since csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}.\newlineCalculation: sin⁡(θ)=12\sin(\theta) = \frac{1}{2}
  3. Find Values of Theta: Step 33: Find the values of θ\theta within the interval −π/2≤θ≤π/2-\pi/2 \leq \theta \leq \pi/2 where sin⁡(θ)=1/2\sin(\theta) = 1/2. Calculation: sin⁡(θ)=1/2\sin(\theta) = 1/2 at θ=π/6\theta = \pi/6 (First Quadrant) Since sin is positive in the first quadrant and negative in the fourth quadrant, and the interval includes only the first quadrant and the negative part of the second quadrant, the only solution in the interval is θ=π/6\theta = \pi/6.

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