# Math

posted by .

the original problem was:

Solve: sin(3x)-sin(x)=cos(2x)

so far i've gotten to:
sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x)

Where would I go from here?

• Math -

you cannot do that!
looks like you factored
sin(3x) - sinx = sinx(sin(2x) - 1)
not true!!!!!!

then sin 90 - sin 30 = sin30(sin60 - 1) Is it???, of course not

we have a formula for sin (3x)
= 3(cos^2 x)(sinx) - sin^3 x
then
sin(3x)-sin(x)=cos(2x)
3(cos^2 x)(sinx) - sin^3 x = 1 - 2sin^2 x
replace cos^2 x with 1 - sin^2x, expand and simplify to get
-4sin^3 x + 2sin^2 x + 2sinx - 1 = 0
2sin^2x(-2sinx + 1) -1(-2sinx + 1) = 0
(-2sinx + 1)(2sin^2x - 1) = 0
sinx = 1/2 or sinx = ± 1/√2

x = 30, 150, 45, 135 degrees

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