solve the equation for solutions over the interval [0, 360)

4cos(squared)x + 4cosx = 1

I'm completely lost on this one!

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- Aug 25th 2008, 10:54 AMneeMore Trigonometric Help Need!
solve the equation for solutions over the interval [0, 360)

4cos(squared)x + 4cosx = 1

I'm completely lost on this one! - Aug 25th 2008, 11:11 AMMoo
Hi !

For you to better understand, substitute $\displaystyle u=\cos(x)$

The equation is now $\displaystyle 4u^2+4u=1$, that is to say $\displaystyle 4u^2+4u-1=0$

Now, use the method of discriminant or complete the square to get the solution :)

A slightly different way to do it is to substitute $\displaystyle u=2 \cos(x)$ ^^ - Aug 25th 2008, 11:35 AMnee
You mean like this?

4cos(squared)x + 4cosx = 1

4cos(squared)x + 4cosx + 1 = 1 + 1

4cos(squared)x + 4cosx +1 = 2

(2cosx + 1)(2cosx + 1) = 2

but then what?

would the following be correct?:

2cosx = 2 or 2cosx = 2 - Aug 25th 2008, 11:41 AMJhevon
no. as Moo said, the equation you are using is $\displaystyle 4 \cos^2 x + 4 \cos x - 1 = 0$

that is what you should be solving. she replaced cos(x) with you to make things look easier. just work on solving the quadratic equation she gave you. you can use the quadratic formula - Aug 25th 2008, 11:47 AMMoo
- Aug 25th 2008, 11:52 AMnee
Ok howbout this then:

cosx = -4 +or- (squareroot)4(squared) - 4(4)(-1) / 2(4)

= -4 +or- (squareroot)0 / 8

= -4 / 8

= -1/2

therefore solution set is:

{120, 240}

???