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\newline−π2≤θ≤π2- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}. Find the value of θ\theta in radians.\newlineθ=−32\theta = -\sqrt{\frac{3}{2}}\newlineWrite your answer in simplified, rationalized form. Do not round.\newlineθ=\theta = ______\newline

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Inverses of sin, cos, and tan: radiansright chevron icon

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Q. \newline−π2≤θ≤π2- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}. Find the value of θ\theta in radians.\newlineθ=−32\theta = -\sqrt{\frac{3}{2}}\newlineWrite your answer in simplified, rationalized form. Do not round.\newlineθ=\theta = ______\newline
  1. Identify angle: Identify the angle where the value of cosine is −3/2-\sqrt{3}/2. In the given interval, the cosine function is negative in the second quadrant. So, we look for an angle in the second quadrant where cos⁡(θ)=−3/2\cos(\theta) = -\sqrt{3}/2. The reference angle for which the cosine value is 3/2\sqrt{3}/2 is π/6\pi/6. Therefore, θ=π−π/6\theta = \pi - \pi/6.
  2. Find quadrant: Calculate the value of θ\theta.θ=π−π6=6π6−π6=5π6\theta = \pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6}.

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