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Question: If \(\cos\theta + \cos 2\theta + \cos 3\theta = 0,\) then the general value of \(\theta\)is...

If cosθ+cos2θ+cos3θ=0,\cos\theta + \cos 2\theta + \cos 3\theta = 0, then the general value of θ\thetais

A

θ=2mπ±2π3\theta = 2m\pi \pm \frac{2\pi}{3}

B

θ=2mπ±π4\theta = 2m\pi \pm \frac{\pi}{4}

C

θ=mπ+(1)m2π3\theta = m\pi + ( - 1)^{m}\frac{2\pi}{3}

D

θ=mπ+(1)mπ3\theta = m\pi + ( - 1)^{m}\frac{\pi}{3}

Answer

θ=2mπ±2π3\theta = 2m\pi \pm \frac{2\pi}{3}

Explanation

Solution

cosθ+cos2θ+cos3θ=0(cosθ+cos3θ)+cos2θ=02cos2θ.cosθ+cos2θ=0cos2θ(2cosθ+1)=0\Rightarrow \cos\theta + \cos 2\theta + \cos 3\theta = 0 \Rightarrow (\cos\theta + \cos 3\theta) + \cos 2\theta = 0 \Rightarrow 2\cos 2\theta.\cos\theta + \cos 2\theta = 0 \Rightarrow \cos 2\theta(2\cos\theta + 1) = 0

cos2θ=0=cosπ2\Rightarrow \cos 2\theta = 0 = \cos\frac{\pi}{2}2θ=2mπ±π/22\theta = 2m\pi \pm \pi/2 θ=mπ±π4\Rightarrow \theta = m\pi \pm \frac{\pi}{4} or

cosθ=12=cos2π3\cos\theta = - \frac{1}{2} = \cos\frac{2\pi}{3} θ=2mπ±2π3.\Rightarrow \theta = 2m\pi \pm \frac{2\pi}{3}.