# Find the general solution (ans +k360 degrees, or as applicable) for `6sin^2x+4sinxcosx-1=0` Please explain. Thank you.

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### 1 Answer

You need to use Pythagorean trigonometric identity `sin^2 x + cos^2 x = 1` to replace the constant term from equation, such that:

`6sin^2 x + 4sin x*cos x - (sin^2 x + cos^2 x) = 0`

`6sin^2 x + 4sin x*cos x - sin^2 x - cos^2 x = 0`

`5sin^2 x + 4sin x*cos x - cos^2 x = 0`

You may solve the homogeneous trigonometric equation by dividing the entire equation by `cos^2 x` , such that:

`5(sin^2 x)/(cos^2 x) + 4(sin x*cos x)/(cos^2 x) -(cos^2 x)/(cos^2 x) = 0`

Reducing duplicate factors yields:

`5tan^2 x + 4tan x - 1 = 0`

You should come up with the following substitution, such that:

`tan x = t`

`5t^2 + 4t - 1 = 0`

Using quadratic formula, yields:

`t_(1,2) = (-4+-sqrt(16 + 20))/10`

`t_(1,2) = (-4+-sqrt36)/10`

`t_1 = 2/10 => t_1 = 1/5`

`t_2 = -10/10 => t_2 = -1`

You need to solve for x the following equations, such that:

`{(tan x = t_1),(tan x = t_2):} => {(tan x = 1/5),(tan x = -1):} => {(x = tan^(-1)(1/5) + k*pi),(x = tan^(-1) (-1) + k*pi):} => {(x = tan^(-1)(1/5) + k*pi),(x = -(pi/4) + k*pi):}`

**Hence, evaluating the general solutions to the given homogeneous trigonometric equation yields **`x = tan^(-1)(1/5) + k*pi; x = -(pi/4) + k*pi.`

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