Question

fnd all solutions on the interval [0,2pi] for the equation 2- sinx= 2cos^2x

Answer 1

2 - sin x = 2 - 2 sin^2x
2sin^2x - sin x = 0
sin x ( 2sinx - 1) = 0

can you continue from here?

Answer 2

i will still be lost from there. what do you do next?

Answer 3

since the product = 0
either sin x = 0 or 2sinx - 1 = 0
so sin x = 0 or sinx = 1/2

sin x = 0 gives x = 0 , pi, 2pi
sin x = 1/2 gives x = pi/6 , 5pi/6

Answer 4

okayy, thank you so much.

Answer 5

no probs