Re: trig problem

From: Joe Blow <justarandomid99@xxxxxxxxx>
Date: Wed, 05 Dec 2007 17:31:41 EST

Hi,

I have been trying to figure these two problems out
and am stumped.
Can anyone help me?

I simply need to solve for x.

-9 sin(3x) - 16 cos(4x) = 0

AND

3 cos(3x) - 4 sin(4x) = 0

Thanks in advance.

Unless my algebra is mistaken, I don't think there are any real solutions. The equations imply

sin(3x) = -16/9*cos(4x)
cos(3x) = 4/3*sin(4x)

Squaring both sides of each equation and adding we have

sin(3x)^2+cos(3x)^2 = 256/81*cos(4x)^2+16/9*sin(4x)^2 = 1

Let cos(4x)^2 = A, then

256/81*A+16/9*(1-A) = 1
==> A = -9/16

So there are no real solutions to your system of equations.
.

Follow-Ups:
- Re: trig problem
  - From: amzoti

References:
- trig problem
  - From: mmcgee00

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