# roots of equationcos^2x=cosx+sin^2xx=?

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The equation to be solved is: (cos x)^2 = cos x + (sin x)^2

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

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

=> 2*(cos x)^2 = cos x + 1

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

=> 2*(cos x)^2 - 2cos x + cos x - 1 = 0

=> 2*(cos x) (cos x - 1) + 1(cos x - 1) = 0

=> (2*(cos x) - 1)(cos x - 1) = 0

=> cos x = 1/2 and cos x = 1

**x = arc cos (1/2) and x = arc cos (1)**

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}.**