# Solve for x cos^2x-cosx-sin^2x=0.

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cos^2 x - cosx - sin^2 x = 0

Let us rearrange terms.

==> cos^2 x -sin^2 x - cosx = 0

But we know that: cos^2 x - sin^2 x = cos2x

==> cos2x - cosx = 0

Now we will use the trigonometric identities to simplify.

We know that cosa - cosb = -2sin(a+b)/2*sin(s-b)2

==> cos2x - cosx = -2sin(3x/2)*sin(x/2).

==> -2sin(3x/2)*sin(x/2) = 0

Then :

sin(3x/2) = 0 ==> 3x/2 = 0, pi, 2pi ==> x = 0, 2pi/3 , 4pi/3

sin(x/2) = 0 ==> x/2 = 0, pi, 2pi ==> x = 0, 2pi, 4p

**==> x = { 0, 2pi/3, 4pi/3 , 2pi}**

We'll recognize the formula for the double angle 2x:

cos 2x = cos (x+x) = cos x*cos x - sin x*sin x

cos 2x = (cos x)^2 - (sin x)^2

We'll write (sin x)^2 = 1 - (cos x)^2 (fundamental formula of trigonometry).

cos 2x = (cos x)^2 - [1 - (cos x)^2]

We'll remove the brackets:

cos 2x = (cos x)^2 - 1 + (cos x)^2

We'll combine like terms:

cos 2x = 2(cos x)^2 - 1 (1)

We'll re-write the equation in cos x, only:

2(cos x)^2 - 1 - cos x = 0

Now , we'll use substitution technique to solve the equation.

We'll note cos x = t and we'll re-write the equation in t:

2t^2 - t - 1 = 0

Since it is a quadratic, we'll apply the quadratic formula:

t1 = {-(-1) + sqrt[(-1)^2 + 4*2*1]}/2*2

t1 = [1+sqrt(1+8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (1-3)/4

t2 = -1/2

Now, we'll put cos x = t1.

cos x = 1

Since it is an elementary equation, we'll apply the formula:

cos x = a

x = +/- arccos a + 2k*pi

x = arccos 1 + 2k*pi

x = 0

x = 2pi

cos x = t2

cos x = -1/2

x = pi - pi/3

x = 2pi/3

x = pi + pi/3

x = 4pi/3

**The solutions for the equation are:{0 ; 2pi/3 ; 4pi/3 ; 2pi}.**