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