How do you solve 6(cos^2)x+cosx=1

Question

I need it in the form that's like (x+x)(x+x) so that I can set them equal to zero and find the answers. Please show all of the steps.

Answer 1

If your expression mean :

6 cos ^ 2 ( x ) + cos ( x ) = 1

then

6 cos ^ 2 ( x ) + cos ( x ) = 1 Subtract 1 to both sides

6 cos ^ 2 ( x ) + cos ( x ) - 1 = 1 - 1

6 cos ^ 2 ( x ) + cos ( x ) - 1 = 0

Substitute :

cos ( x ) = u

6 u ^ 2 + u - 1 = 0

The exact solution are :

u = - 1 / 2

u = 1 / 3

cos ( x ) = - 1 / 2

cos ( x ) = 1 / 3

Cosine are periodic functions , with period 2 pi

Solutions :

[ cos ^ - 1 ( - 1 / 2 ) = 120 ° = 2 pi / 3 rad ]

x = 2 pi n + 2 pi / 3

x = 2 pi n - 2 pi / 3

AND

x = 2 pi n + cos ^ - 1 ( 1 / 3 )

x = 2 pi n - cos ^ - 1 ( 1 / 3 )

Remark :

cos ^ - 1

is the inverse cosine function

P.S

If you don't know how to solve equation :

6 u ^ 2 + u - 1 = 0

In google type:

quadratic equation online

When you see list of results click on:

Free Online Quadratic Equation Solver:Solve by Quadratic Formula

When page be open in rectangle type:

6 u ^ 2 + u - 1 = 0

and click option: solve it!

You wil see solution step-by-step