# Thread: More Trigonometric Help Need!

1. ## More 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!

2. Hi !
Originally Posted by nee
solve the equation for solutions over the interval [0, 360)

4cos(squared)x + 4cosx = 1

I'm completely lost on this one!
For you to better understand, substitute $u=\cos(x)$

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

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

A slightly different way to do it is to substitute $u=2 \cos(x)$ ^^

3. 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

4. Originally Posted by nee
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
no. as Moo said, the equation you are using is $4 \cos^2 x + 4 \cos x - 1 = 0$

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

5. Originally Posted by nee
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
Eh !!!
That's good ! It's just like completing the square

(2cos(x)+1)(2cos(x)+1)=(2cos(x)+1)²=2
Thus $2 \cos(x)+1=\pm \sqrt{2}$
And go from here

6. 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}

???