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.
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