Solve the equation $\cos 2x = \cos x$ for $0 \le x \le 2\pi$ .

x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}

only

x = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, 2\pi

x = 0, \tfrac{\pi}{3}, \tfrac{5\pi}{3}, 2\pi

x = 0, \pi, 2\pi

Explanation:

2\cos^2x-\cos x-1=0 \Rightarrow (2\cos x+1)(\cos x-1)=0

. So

\cos x=1

giving

x=0,2\pi

, or

\cos x=-\tfrac12

giving

x=\tfrac{2\pi}{3},\tfrac{4\pi}{3}

Derived from ZIMSEC Mathematics Paper 1, June 2011, Q11

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