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