# Thread: problems with cosine equation

1. ## problems with cosine equation

i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!

2. Originally Posted by airportman92
i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!
nevermind, i got it i think!

3. Originally Posted by airportman92
i have 2cos(2x)-2cos(x)=0 and im trying to find the values of x. i already figured out that they are +/- 2pi/3 but i dont know how to work that out showing work. all i have so far is cos2x=cosx but i dont know what to do with that. any help at all would be appreciated since i do not know how to approach this, thank u!

Note that $\displaystyle \cos\!\left(2x\right)=2\cos^2x-1$.

So we get $\displaystyle 2\cos^2x-1-\cos x=0$

Note that this is quadratic in $\displaystyle \cos x$. So letting $\displaystyle u=\cos x$, we get $\displaystyle 2u^2-u-1=0$.

I leave it for you to solve for u (you'll get 2 different values). Then set $\displaystyle \cos x$ equal to the values and solve for x. Does this make sense?

4. .....sorry.....
wrong post

5. Hi there airportman92

$\displaystyle 2\cos(2x)-2\cos(x)=0$

needs a substitution $\displaystyle \cos(2x)=2\cos^2(x)-1$

Therefore you have $\displaystyle 2(2\cos^2(x)-1)-2\cos(x)=0$

Now expanding gives you

$\displaystyle 4\cos^2(x)-2-2\cos(x)=0$

dividing through by 2 gives

$\displaystyle 2\cos^2(x)-1-\cos(x)=0$

and re arranging

$\displaystyle 2\cos^2(x)-\cos(x)-1=0$

now making $\displaystyle a = \cos(x)$

$\displaystyle 2a^2-a-1=0$

6. ooo ok thanks!

what about its derivative which i got to be -4sin2x+2sinx.

7. correct!

$\displaystyle (2\cos(2x)-2\cos(x))'= -4\sin(2x)+2\sin(x)$

8. Originally Posted by pickslides
correct!

$\displaystyle (2\cos(2x)-2\cos(x))'= -4\sin(2x)+2\sin(x)$
but the x-values are impossible to solve for. i know its supposed to be cos^-1(.25) but theres no way to show that

9. Originally Posted by airportman92
but the x-values are impossible to solve for. i know its supposed to be cos^-1(.25) but theres no way to show that

Are you talking about the original question or something to do with the derivative?

10. Originally Posted by pickslides
Are you talking about the original question or something to do with the derivative?

well this is already the derivative of somethin, so i just neeed the x value which i thought i got

11. Originally Posted by airportman92
well this is already the derivative of somethin, so i just neeed the x value which i thought i got

Did you want to solve for x with respect to the original equation or the derivative? 2 different things!