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

1 Answer
Noah G
Mar 19, 2018

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!

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