Solve Cos2x=cosx

Answer: \(x = 2\pi k\) or \(x = \frac{2\pi}{3} + 2\pi k\) or \(x = \frac{4\pi}{3} + 2\pi k\), for any integer \(k\).

Explanation:

Steps:

  1. Use the double-angle identity: \(\cos 2x = 2\cos^2 x - 1\).
  2. Set equal to \(\cos x\): \(2\cos^2 x - 1 = \cos x\).
  3. Let \(u=\cos x\). Solve the quadratic \(2u^2 - u - 1 = 0\):

\[ u = \frac{1 \pm \sqrt{1+8}}{4} = \frac{1\pm3}{4}, \]

so \(u=1\) or \(u=-\tfrac{1}{2}\).

  1. If \(\cos x = 1\) then \(x = 2\pi k\).

If \(\cos x = -\tfrac{1}{2}\) then \(x = \frac{2\pi}{3} + 2\pi k\) or \(x = \frac{4\pi}{3} + 2\pi k\).

All solutions given with \(k\in\mathbb{Z}\).