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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- August 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! - August 25th 2008, 11:11 AMMoo
Hi !

For you to better understand, substitute

The equation is now , that is to say

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

A slightly different way to do it is to substitute ^^ - August 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 - August 25th 2008, 11:41 AMJhevon
- August 25th 2008, 11:47 AMMoo
- August 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}

???