Solve sin5x-sin3x+sinx = 0

solve sin5x-sin3x+sinx = 0, --> means finding values of x
for 0deg (< or =) x (< or =) 180deg

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i dont know how to type the symbol with a < and a _ below it here, sorry.
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answers: 0deg, 60deg, 120deg, 180deg, 30deg, 150deg


help, thanks.

sinP - sinQ = 2cos½(P+Q)sin½(P-Q)
therefore sin5x-sin3x=2cos4xsinx
sin5x - sin3x +sinx = 2cos4xsinx + sinx = 0
sinx(2cos4x+1) = 0

sinx = 0
x=0,180

OR

2cos4x+1=0
cos4x = -½
if 0 4x = 120, 240, 480, 600 (from ASTC quadrants)
x = 30, 60, 120, 150

therefore x = 0, 30, 60, 120, 150, 180