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Exercise 3.1 · Q5

Q.Find the principal solutions of the following equation: sin⁡θ=−12\sin\theta = -\frac{1}{2}

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sin⁡θ=−12\sin\theta=-\dfrac12 is negative, so θ\theta is in the third or fourth quadrant, with reference angle π6\dfrac{\pi}{6} (since sin⁡π6=12\sin\dfrac{\pi}{6}=\dfrac12). Third quadrant: θ=π+π6=7π6\theta=\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6}. Fourth quadrant: $\theta=2\pi-\df …

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