# 4cos²xsinx-2sin²x=3sinxsolve for x

2
by aswintrichy

2014-08-10T18:41:34+05:30

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4 cos² x  sin x  - 2 sin² x = 3 sin x

if sinx is 0 then  x = 2 n π  , n = natural number

if sinx is not 0 then, cancel sin x on both sides,

4 cos² x - 2 sin x = 3
4 - 4 sin² x - 2 sin x - 3 =0
4 sin² x + 2 sin x -1 = 0
Δ = 4 + 16 = 20
sin x = (-2 +- 2√5) / 8    =  (-1 +- √5) / 4  =
x =  2 n π + 18 deg  or  2 n π (- 54 deg  = 126 deg)

2014-08-10T18:49:26+05:30
4cos²xsinx-2sin²x = 3sinx
4(1-sin²x)sinx-2sin²x = 3sinx
4sinx-4sin³x-2sin²x = 3sinx
4sin³x+2sin²x-sinx = 0
sinx(4sin²x+2sinx-1) = 0
sinx = 0 or 4sin²x+2sinx-1 = 0
x = 0 or sinx = {-2+-√(2²-4*4*(-1))}/2*4
x = 0 or sinx = (-1+-√5)/4
x = 0 or sinx = (-1-√5)/4  or sinx = (-1+√5)/4
x = 0 or x = sin^-1{(-1-√5)/4}  or  x = sin^-1{(-1+√5)/4}
there is no boundary condition on x so general solution of x is
x = (nπ+(-1)^n 0) or x = [nπ+(-1)^n sin^-1{(-1-√5)/4}] or                    (where n is an integer)
x = [nπ+(-1)^n sin^-1{(-1+√5)/4}]
hence
x = nπ  or
x = [nπ+(-1)^n sin^-1{(-1-√5)/4}]   or
x = [nπ+(-1)^n sin^-1{(-1+√5)/4}]