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Exercise 3.1 · Q1

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

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✓ Free question

cos⁡π3=12\cos\dfrac{\pi}{3}=\dfrac12 and π3∈[0,2π)\dfrac{\pi}{3}\in[0,2\pi), so π3\dfrac{\pi}{3} is a principal solution. Using cos⁡θ=cos⁡(2π−θ)\cos\theta=\cos(2\pi-\theta): cos⁡(2π−π3)=cos⁡π3=12\cos\left(2\pi-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{3}=\dfrac12, i.e. cos⁡5π3=12\cos\dfrac{5\pi}{3}=\dfrac12, and 5π3∈[0,2π)\dfrac{5\pi}{3}\in[0,2\pi) too.

✓Final answer

θ=π3, 5π3\theta=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}

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