Solve `3cosx+3=2sin^2x` for x in `(0,2pi)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation `3cosx+3=2sin^2x` has to be solved for x in `(0, 2*pi)`

`3cosx+3=2sin^2x`

=> `2*sin^2x - 3*cos x - 3 = 0`

=> `2*(1 - cos^2x) - 3*cos x - 3 = 0`

=> `2 - 2*cos^2x - 3*cos x - 3 = 0`

=> `2*cos^2x + 3*cos x + 1 = 0`

=> `2*cos^2x + 2*cosx + cos x + 1 = 0`

=> `2*cosx(cos x + 1) + 1(cos x + 1) = 0`

=> `(2*cos x + 1)(cos x + 1) = 0`

=> `cos x = -1/2` and `cos x = -1`

For x in `(0, 2*pi)` , x = 120, x = 240 and x = 180 degrees

The solution of the equation is x = 120, x = 240 and x = 180 degrees.

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