Question

# Find all values of $$\theta$$ between $${0}^{o}$$ and $${180}^{o}$$ satisfying the equation : $$\cos{6\theta}+\cos{4\theta}+\cos{2\theta}+1=0$$

Solution

## $$cos 6 \theta + cos 4 \theta + cos 2 \theta + 1 = 0$$$$\Rightarrow 2 cos \, 4 \theta \, cos \, 2 \theta + cos 4 \theta + 1 = 0$$$$\Rightarrow 2 cos 4 \theta \, cos 2 \theta + 2 cos^2 2 \theta = 0$$$$\Rightarrow 2 cos 2 \theta \, (cos 4 \theta + cos 2 \theta) = 0$$$$\Rightarrow 2 cos 2 \theta \times 2 cos 3 \theta \times cos \theta = 0$$$$\therefore cos \theta = 0, \, cos 2 \theta = 0, \, cos 3 \theta = 0$$$$\theta = (2n + 1) \dfrac{\pi}{2}$$ or $$\theta = (2n + 1) \dfrac{\pi}{4}$$ or $$\theta = (2n + 1)$$Mathematics

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