How can I solve for x?

Question

I am trying to work out a calc problem and am stuck at solving for x. Right now I have
0=2cosx+2cos(2x)

I need to know what value for x results in 0; please explain how to do this.

Answer 1

2cos(x) + 2cos(2x) = 0
cos(x) + cos(2x) = 0

Remember that cos(2x) = 2cos²(x) - 1
Substitute this back into the equation:
cos(x) + 2cos²(x) - 1 = 0
2cos²(x) + cos(x) - 1 = 0
2cos²(x) - cos(x) + 2cos(x) - 1 = 0
cos(x)(2cos(x) - 1) + 1(2cos(x) - 1) = 0
(cos(x) + 1)(2cos(x) - 1) = 0
cos(x) = {-1, 1/2}

x = (2n+1)π
OR
x = 2nπ ± π/3

Answer 2

The first trick to this equation is to realise that
Cos 2x = Cos^2 x - Sin^2 x
We already have something with a Cos x in it, so it might be a good idea to rewrite everything in terms of Cos x, but we have this nasty Sin x here... Luckily we can use
Sin^2 x = 1 - Cos^2 x.
giving us
4Cos^2 x + 2Cos x - 2
The 2nd trick to this question is to notice that this is just a quadratic in Cos x, so lets use the substitution
y = Cos x
to make this clearer
2y^2 + y - 1 = 0
now we just solve this quadratic
y = -1, 1/2
And now all thats left to do is find x.
I will leave the rest to you, making sure that you get your quadrants right.

Answer 3

cos 2x + cos x = 0
2 cos ² x + cos x - 1 = 0
(2 cos x - 1) (cos x + 1) = 0
cos x = 1/2 , cos x = - 1
x = π/3 , 5π/3 , π

Answer 4

2cosx = -2cos(2x)
Let cosx = q
2cosx = 2q
-2cos(2x) = 2q
cos(2x) = -q
If x=pi/3, cosx = sqrt(3)/2 and 2cosx = sqrt(3)
2x = 2pi/3, cos(2x) = -sqrt(3)/2, and -2cos(2x)=sqrt(3)