Question

How do you find the general solutions for `6sin x cos x +3sin x = 2cos x + 1`?

Answer 1

Put all terms to one side of the equation .

`6sinxcosx - 2cosx + 3sinx - 1 = 0`

`2cosx(3sinx - 1) + (3sinx - 1) = 0`

`(2cosx + 1)(3sinx - 1) = 0`

`cosx = -1/2 or sinx = 1/3`

`x = 120˚ + 360˚n, 240˚ + 360˚n, arcsin(1/3) + 360˚n, 180˚ - arcsin(1/3) + 360˚n`

Hopefully this helps!