Solve sin(3θ) - sin(2θ) = sin(6θ)sin(θ) for θ is less than 180 but greater than 0

I've got as far as:

2cos(2θ)(sin(θ) - sin(4θ)) = 0

And I'm now stuck.

Any help would be great :)

Printable View

- Jan 21st 2013, 08:47 AMkinhew93Trigonometric equation
Solve sin(3θ) - sin(2θ) = sin(6θ)sin(θ) for θ is less than 180 but greater than 0

I've got as far as:

2cos(2θ)(sin(θ) - sin(4θ)) = 0

And I'm now stuck.

Any help would be great :) - Jan 21st 2013, 07:34 PMchiroRe: Trigonometric equation
Hey kinhew93.

If your new expression is correct then cos(2x) = 0 and sin(x) = sin(4x). The cos term is easy and the sin term is also good when sin(2x) = 2sin(x)cos(x) which means you will get sin(4x) = 2*2*sin(x)*cos(x)*cos(2x) = sin(x) which means 4cos(x)*cos(2x) = 1 or cos(x)*cos(2x) = 1/4 which can be further reduced.