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Miscellaneous Exercise 3 · Q89

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

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As in II.1A, sin⁡2θ=−12\sin2\theta=-\dfrac12 over the extended range [0,4π)[0,4\pi) for 2θ2\theta gives four solutions after dividing by 2: θ=7π12,11π12,19π12,23π12\theta=\dfrac{7\pi}{12},\dfrac{11\pi}{12},\dfrac{19\pi}{12},\dfrac{23\pi}{12}. …

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