How solve equation? (equation in sin angle, cos angle)cos3x-10sin^2x=3cosx-10

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to write the equation in terms of one function, `sin x`  or `cos x` .

I suggest you to write the equation in terms of `cos x`  because if you decide to work in terms of `sin x,`  the original equation turns into a difficult equation (you need to write `cos x = sqrt(1 - sin^2 x))` .

You need to write the equation in terms of `cos x` , hence you need to write `cos 3x = cos(2x + x).`

You need to expand `cos (2x + x) = cos 2x*cos x - sin 2x*sin x`

`cos (2x + x) = (2cos^2 x - 1)*cos x - 2sin^2 x*cos x`

`cos (2x + x) = 2cos^3 x - cos x - 2cos x + 2cos^3 x`

`cos (2x + x) = 4cos^3 x - 3cos x`

Writing the equation in terms of `cos x`  yields:

`4cos^3 x - 3cos x - 10(1 - cos^2 x) = 3cos x - 10`

Opening the brackets yields:

`4cos^3 x - 3cos x - 10 + 10cos^2 x = 3cos x - 10`

Reducing the opposite terms and moving the terms to the left side yields:

`4cos^3 x - 3cos x + 10cos^2 x- 3cos x = 0`

`4cos^3 x + 10cos^2 x - 6cos x = 0`

Factoring out `2cos x`  yields:

`2cos x*(2cos^2 x + 5cos x - 3) = 0`

`2cos x = 0 =gt cos x = 0 =gt x = +-pi/2 + 2n*pi`

`2cos^2 x + 5cos x - 3 = 0`

You should come up with the substitution cos x = y, hence:

`2y^2 + 5y - 3 = 0`

Using quadratic formula yields:

`y_(1,2) = (-5 +- sqrt(25 + 24))/4 =gt y_1 = (-5+7)/4 =gt y_1 = 1/2`

`` `y_2 = (-5-7)/4 =gt y_2 = -3`

You need to find x, hence `cos x = y_1 =gt cos x = 1/2`

`x = +-pi/3 + 2n*pi`

Equating `cos x = y_2`  yields a contradiction since `|cos x|=lt1.`

Hence, the solutions to the equation are `x_1 = +-pi/2 + 2n*pi; x_2 = +-pi/3 + 2n*pi` .

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question