Question

How do you solve `1-sin2x = cosx- sinx`?

Answer 1

Arc x = 0; `x = pi/4; x = pi/2; x = pi`

Explanation:

1 - sin 2x = cos x - sin x
Square both sides:
`(1 - sin 2x)^2 = (cos x - sin x)^2`
`1 - 2sin 2x + sin^2 2x = cos^2 x + sin^2 x - 2sin x.cos x`
`1 - 2sin 2x + sin^2 2x = 1 - sin 2x`
`sin^2 2x - sin 2x = 0`
sin 2x (sin 2x - 1) = 0
a. sin 2x = 0
sin 2x= 0 --> 2x = 0 --> x = 0
sin 2x = 0 --> `2x = pi`--> `x = pi/2`
sin 2x = 0 --> `2x = 2pi` --> `x = pi`
b. sin 2x = 1 --> `2x = pi/2` --> `x = pi/4`