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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.\newlinesin⁡(θ)=0\sin(\theta)=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.\newlinesin⁡(θ)=0\sin(\theta)=0
  1. Identify Zero Points: We need to identify the angles for which the sine function is equal to zero within the given interval. The sine function is equal to zero at specific points on the unit circle, namely where the yy-coordinate is zero. This occurs at the angles where the corresponding point on the unit circle intersects the xx-axis.
  2. Check Interval: Within the interval −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, the sine function is zero at θ=0\theta = 0, because sin⁡(0)=0\sin(0) = 0. We need to check if there are any other points within the interval where sin⁡(θ)\sin(\theta) could be zero.
  3. Zero at Multiples of π\pi: The sine function is also zero at multiples of π\pi, but within our interval, the only multiple of π\pi is 00 itself. Therefore, there are no other solutions within the interval −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.
  4. Conclusion: We conclude that the only solution to sin⁡(θ)=0\sin(\theta) = 0 within the interval −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} is θ=0\theta = 0.

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