Expert: Sherman D. - 3/12/2008

Question
hi i'm having a real hard time solving this problem

sin(3x)+ cos(2x)=0 on the scale 0-2pi

Answer
sin(3x) + cos(2x) = 0
sin(2x + x) + cos(2x) = 0
sin(2x)cos(x) + sin(x)cos(2x) + cos(2x) = 0
2sin(x)cos(x)^2 + sin(x)(2cos(x)^2 - 1) + 2cos(x)^2 - 1 = 0
2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x) + 2cos(x)^2 - 1 = 0
4sin(x)cos(x)^2 - sin(x) + 2cos(x)^2 - 1 = 0
4sin(x)(1 - sin(x)^2) - sin(x) + 2(1 - sin(x)^2) - 1 = 0
4sin(x) - 4sin(x)^3 - sin(x) + 2 - 2sin(x)^2 - 1 = 0
-4sin(x)^3 - 2sin(x)^2 + 3sin(x) + 1 = 0
(-sin(x))(4sin(x)^2 + 2sin(x) - 3) = -1

thats as far as i got. If you know how to work the cubic formula, then your all good.

www.quickmath.com gives the answers of

x = (3pi/2), (3pi/10), and (7pi/10)