# math

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find all solutions in the interval [0,2 pi)

sin(x+(3.14/3) + sin(x- 3.14/3) =1

sin^4 x cos^2 x

Since sin (a+b) = sina cosb + cosb sina
and
sin (a-b) = sina cosb - cosb sina,
the first problem can be written
2 sin x cos (pi/3)= sin x
The solution to sin x = 1 is x = pi/2

sin^4 cos^2 x = sin^4 x(1 - sin^2x)=0
The solutions are sin x = 0 and sin^2 x = 1. That would correspond to x=0, pi/2, pi, and 3 pi/4

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