Question

Find the general solution of the following .

2sin²x+3cosx=0

Answer 1

Use the main Pythagorean trigonometric identity to get everything in terms of one trig function (cosine):
sin²θ + cos²θ = 1
sin²θ = 1 - cos²θ
- - - - - - - - - - - - - - - -

2·sin²(x) + 3·cos(x) = 0
2·[ 1 - cos²(x) ] + 3·cos(x) = 0
2 - 2·cos²(x) + 3·cos(x) = 0
2·cos²(x) - 3·cos(x) - 2 = 0

Factor the trinomial (or use the quadratic formula):
2·cos²(x) - 4·cos(x) + cos(x) - 2 = 0
2·cos(x) · [ cos(x) - 2 ] + 1 · [ cos(x) - 2 ] = 0
[ 2·cos(x) + 1 ] · [ cos(x) - 2 ] = 0

2·cos(x) + 1 = 0
or
cos(x) - 2 = 0

Which gives:
cos(x) = -½
or
cos(x) = 2

Since cosine cannot be greater than one, it cannot be two.

Cosine is -½ when:

x=120 or x=240

Therefore general solution can be written as

x=120360n

where n is any integer

If you want that in radians:
x = 2π/3 2πn

where n is any integer