SOLUTION: 2sin^2x+7cosx-5=0 This is as far as I've gotten, not sure how to finalize the answer... 2(1-cos^2x)+7cosx-5=0 2-2cos^2x+7cosx-5=0 -1(-2cos^2x+7cosx-3)=0(-1) 2cos^2x-7cosx+3=0

Algebra ->  Trigonometry-basics -> SOLUTION: 2sin^2x+7cosx-5=0 This is as far as I've gotten, not sure how to finalize the answer... 2(1-cos^2x)+7cosx-5=0 2-2cos^2x+7cosx-5=0 -1(-2cos^2x+7cosx-3)=0(-1) 2cos^2x-7cosx+3=0       Log On


   

Question 961977: 2sin^2x+7cosx-5=0
This is as far as I've gotten, not sure how to finalize the answer...
2(1-cos^2x)+7cosx-5=0
2-2cos^2x+7cosx-5=0
-1(-2cos^2x+7cosx-3)=0(-1)
2cos^2x-7cosx+3=0
factor to...
(2cosx-1)(cosx-3)
cosx= 1/2-----on unit circle...?
cosx= 3------- on unit circle?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
For[0,2π)
2sin^2x+7cosx-5=0
2(1-cos^2x)+7cosx-5=0
2-2cos^2x+7cosx-5=0
2cos^2x-7cosx+3=0
(2cosx-1)(cosx-3)
2cosx-1=0
cosx=1/2
x=π/3, 5π/3
or
cosx-3=0
cpsx=3 (reject,( -1 ≤ cosx ≤ 1))