lgg23 5 years ago Solve the following trig equation for all x in the interval [0,2pi): 2 [cos x]^(2) + cos x sin x = 1

1. nowhereman

use double-arc formulas

2. anonymous

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.

3. anonymous

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