Posted by **Katie** on Thursday, April 23, 2009 at 7:17pm.

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 -
**Reiny**, Friday, April 24, 2009 at 12:49am
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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