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Question: The general solution of the equation $\sin 2\theta = \frac{\sqrt{5}-1}{4}$ is...

The general solution of the equation sin⁡2θ=5−14\sin 2\theta = \frac{\sqrt{5}-1}{4} is

A

θ=nπ2+(−1)nπ20,n∈Z\theta = \frac{n\pi}{2} + (-1)^n \frac{\pi}{20}, n \in Z

B

θ=nπ+(−1)nπ5,n∈Z\theta = n\pi + (-1)^n \frac{\pi}{5}, n \in Z

C

θ=2nπ±π10,n∈Z\theta = 2n\pi \pm \frac{\pi}{10}, n \in Z

D

θ=nπ±π10,n∈Z\theta = n\pi \pm \frac{\pi}{10}, n \in Z

Answer

(a)

Explanation

Solution

Since (√5 – 1)/4 = sin 18° = sin(π/10), we have 2θ = π/10 + 2πn or 2θ = π – (π/10) + 2πn. Dividing by 2 gives θ = π/20 + nπ or θ = 9π/20 + nπ. This is neatly written as θ = (nπ)/2 + (–1)ⁿ (π/20).