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Miscellaneous Exercise 3 · Q86

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

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Let ϕ=2θ\phi=2\theta, so ϕ\phi ranges over [0,4π)[0,4\pi) as θ\theta ranges over [0,2π)[0,2\pi). sin⁡ϕ=−12\sin\phi=-\dfrac12 gives ϕ=7π6,11π6\phi=\dfrac{7\pi}{6},\dfrac{11\pi}{6} in the first 2π2\pi-cycle, and 7π6+2π=19π6,11π6+2π=23π6\dfrac{7\pi}{6}+2\pi=\dfrac{19\pi}{6},\dfrac{11\pi}{6}+2\pi=\dfrac{23\pi}{6} in the second. Dividing all four by 22: $\theta=\dfrac{7\pi}{12},\df …

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