sin 2x = cos3x = sin(pi/2 - 3x)
Therefore, one solution is
2x = pi/2 - 3x
5x = pi/2
x = pi/10 = 18 degrees
There is a second solution at x = pi/2, but that is just outside the domain of interest.
Do you mean
sin^2 x = cos^3 x
or do you mean
sin(2x) = cos(3x)
If you mean sin(2x) = cos(3x)
2 sin x cos x = 4 cos^3 x - 3 cos x
2 sin x = 4 cos^2 x - 3
2 sin x = 4 (1 - sin^2 x) - 3
2 sin x = 4 - 4 sin^2 x - 3
4 sin^2 x + 2 sin x -1 = 0
let z = sin x
4 z^2 + 2 z - 1 = 0
z = [ -2 +/- sqrt (4+16) ] / 8
z = [ -2 +/- 2 sqrt 5 ] /8
sin x = (-1 +/- sqrt 5)/4
if x in quadrant 1 then use + root
sin x = (sqrt 5 - 1)/4
Do it the way Dr WLS did it, much simpler and more likely correct.
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