citlalivazquez
1

# 2cos^2(x)-5cos(x)+2=0

SHAPARISJOHNSON

2cos^2(x) - 5cos(x) + 2 = 0 Think of "cox(x)" as just x for now, just like any other variable. 2x^2 - 5x + 2 = 0 Now factor it, just like it's a quadratic equation. (2x - 1) (x - 2) = 0 2x - 1 = 0 x - 2 = 0 x = 1/2 or x = 2 Now change the variable x, back to "cos(x)", just the way it started out as. cos(x) = 1/2 or cos(x) = 2 Now use a unit circle or graphing calculator to find the angle measures that make this true.

lyssie11234

$2cos^2x-5cos+2=0\\\\cosx=t\in < -1;\ 1 >\\\\2t^2-5t+2=0\\\\a=2;\ b=-5;\ c=2\\\\\Delta=b^2-4ac;\ \Delta=(-5)^2-4\cdot2\cdot2=25-16=9\\\\t_1=\frac{-b-\sqrt\Delta}{2a};\ t_2=\frac{-b+\sqrt\Delta}{2a}\\\\t_1=\frac{5-\sqrt9}{2\cdot2}=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\in < -1;\ 1 >\\\\t_2=\frac{5+\sqrt9}{2\cdot2}=\frac{5+3}{4}=\frac{8}{4}=2\notin < -1;\ 1 >$ $cosx=\frac{1}{2}\to x=\frac{\pi}{3}+2k\pi\ \vee\ x=-\frac{\pi}{3}+2k\pi\ \ \ (k\in\mathbb{Z})$