\cos\theta = \sqrt{3}

February 22, 2026

$\cos\theta = \sqrt{3}$

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\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = 2 \cos\theta \cos 60^\circ
Since $\cos 60^\circ = \frac{1}{2}$, we have:
\cos\theta \cdot \frac{1}{2} = \cos\theta
But $\cos\theta = \sqrt{3}$ has no real solution since $|\cos\theta| \leq 1$. This suggests an error in assumption—we must re-evaluate the expression. Wait: correction in identity use. Actually:
\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = 2 \cos\theta \cos 60^\circ = 2 \cos\theta \cdot \frac{1}{2} = \cos\theta
So the equation is $\cos\theta = \sqrt{3}$, which is impossible. Thus, reconsider the original problem—instead, suppose the equation was meant to be:

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