# maths

posted by
**elle**
.

hey, i would really appreciate some help solving for x when:

sin2x=cosx

Use the identity sin 2A = 2sinAcosA

so:

sin 2x = cos x

2sinxcosx - cosx = 0

cosx(2sinx - 1)=0

cosx = 0 or 2sinx=1, yielding sinx=1/2

from cosx=0 and by looking at the cosine graph, we conclude that x=pi/2 or 3pi/2 (90º and 270º)

from sinx = 1/2 and knowing that the sine is positive in the first and second quadrants, we conclude that x = pi/6 or 5pi/6 (30º and 150º)

those are the solutions in the domain 0 ≤ x ≤ 2pi,

if you want a general solution, add 2(k)pi to each of the four solutions I gave you, where k is an integer.