Thread: trigonometric equation

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

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.

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)?

oh no w8 solved it

squared it wrong