Question

2sin^2 + 5cosx=4 , solve for x

Answer 1

Replace sin^2x with 1-cos^2x

Answer 2

^^ after this solve it as a quadratic equation, you could say u = cos x
then
2u^2 -5u +2 = 0

Answer 3

isnt it 2x^2 -5x-3=0

Answer 4

Mertsj suggests replace sin^2 = 1 -cos ^2 . So you have 2 sin^2 = 2 -2cos^2 . Replace to your equation you have 2 -2cos^2 + 5cos =4
---> -2cos^2 +5cos -2 =0 or cos^2 -5cos +2 =0. let u = cos . you have
2u^2 -5u +2=0
don't know how you get 3. Recheck.