I have a quiz tomorrow but i don't really get it:
1. Given 4cos x-2=cos x, solve for x, to the nearest degree, in the interval 0 < x < 360
i got 3cosx-2=0 but i don't know what to do after that..
and
2. Solve 2cos$\displaystyle ^2$θ-cosθ=1
I have a quiz tomorrow but i don't really get it:
1. Given 4cos x-2=cos x, solve for x, to the nearest degree, in the interval 0 < x < 360
i got 3cosx-2=0 but i don't know what to do after that..
and
2. Solve 2cos$\displaystyle ^2$θ-cosθ=1
(1) 4 cos x - cos x = 2
3 cos x = 2
cos x = 2/3
basic angle x = 48.19
Note that cos x is positive , thus it can be either in the first quadrant or the fourth quadrant .
therefore , x = 48
or
x = 360 - 48.19
=311.81
= 311
(2) 2 cos^2 x - cos x -1 = 0
( 2 cos x + 1 ) ( cos x - 1 ) = 0
cos x = - 1/2 or cos = 1
use the same method to solve .