Find all the zeros of this function: f(x) = 1 - 3cos(2x) in the interval (pi/2 is less than/equal to x less then equal to pi.
Thank you.
Find all the zeros of this function: f(x) = 1 - 3cos(2x) in the interval (pi/2 is less than/equal to x less then equal to pi.
Thank you.
$\displaystyle \frac{\pi}{2} < x < \pi$Quote:
Originally Posted by ADH
$\displaystyle \pi < 2x < 2\pi$
let $\displaystyle u = 2x$ ...
$\displaystyle 1 - 3\cos(u) = 0$
$\displaystyle \cos(u) = \frac{1}{3}$
$\displaystyle u$ is in quad IV , so ...
$\displaystyle u = 2\pi - \arccos\left(\frac{1}{3}\right)$
$\displaystyle 2x = 2\pi - \arccos\left(\frac{1}{3}\right)$
$\displaystyle x = \pi - \frac{1}{2}\arccos\left(\frac{1}{3}\right)$
