# Math Help - trigonometric equation

1. ## trigonometric equation

How do I start in solving 6sinx = 5 - 8cosx????

2. Originally Posted by xwrathbringerx
How do I start in solving 6sinx = 5 - 8cosx????
One way is you use sin(x) = sqrt[1 -cos^2(x)].
Another way is you use cos(x) = sqrt[1 -sin^2(x)].

Or, just square both sides of the posted equation.

The idea is to arrive to an equation that involves one trig function only.

3. One way is you use sin(x) = sqrt[1 -cos^2(x)].
Another way is you use cos(x) = sqrt[1 -sin^2(x)].

Or, just square both sides of the posted equation.
I've tried all three but i cannot get the answer out of them (the answers are meant to be 97 degrees and 337 degrees)?

4. oh no w8 solved it

squared it wrong