Question

cos2x=1/2 interval (0, pi) solve the equation

Answer 1

$\cos (2x) = \frac{1}{2}$ interval $[0,\pi ]$ solve the equation

Solution.

Angles of the unit circumference which have cosine equal to $\frac{1}{2}$ : $\alpha + 2\pi k$ and $-\alpha + 2\pi k$ , where $\alpha = \arccos \left(\frac{1}{2}\right) = \frac{\pi}{3}$ ; $k$ is integer. We can write this in the form: $\alpha = \pm \arccos \left(\frac{1}{2}\right) + 2\pi k = \pm \frac{\pi}{3} + 2\pi k$ . Comparing with the original equation we have $2x = \alpha$ ; $x = \frac{\alpha}{2} = \pm \frac{\pi}{6} + \pi k$ . There is only one solution in the interval $[0, \pi]$ : $x = \frac{\pi}{6}$ .

Answer: $x = \frac{\pi}{6}$ .