Question

Solve the following trig equation for all x in the interval [0,2pi):

2 [cos x]^(2) + cos x sin x = 1

Answer 1

use double-arc formulas

Answer 2

not sure I understand the question. so I'm not sure if this is the answer.\[(2\cos ^{2}x-1)+cosxsinx=0;2((\cos2x)+cosxsinx=0));2\cos2x+2sinxcosx=0;2\cos2x+\sin2x=0\] if you put in 0 for x, you get 1. If you put \[\pi\] for x, you get 1also. I hope this is the answer you need.

Answer 3

apply the equation (cos x)^2 +(sin x)^2=1,then get (sin x)^2=(sin x)(cos x)