# What is the solution to cos^2x - sin^2x + sin x = 0.

*print*Print*list*Cite

### 2 Answers

We have to solve: (cos x)^2 - (sin x)^2 + sin x = 0

(cos x)^2 - (sin x)^2 + sin x = 0

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

=> 1 - (sin x)^2 - (sin x)^2 + sin x = 0

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

let sin x = y

=> 1 - 2*y^2 + y = 0

=> 2y^2 - y - 1 = 0

=> 2y^2 - 2y + y - 1 = 0

=> 2y(y - 1) + 1(y - 1) = 0

=> (2y + 1)(y - 1) = 0

2y + 1 = 0

=> y = -1/2

=> sin x = -1/2

=> x = arc sin (-1/2)

=> x = -30 + n*360 degrees

y - 1 = 0

=> y = 1

=> sin x = 1

=> x = arc sin (1)

=> x = 90 + n*360

**The solution of the equation are x = -30 + n*360 degrees and x = 90 + n*360**

Since there are 2 terms that contain the fuction sine, we'll replace (cos x)^2, by the difference 1 - (sin x)^2 (from Pythagorean identity).

The equation will become:

1 - 2(sin x)^2 + sin x = 0

We'll multiply by -1 and we'll re-arrange the terms:

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

We'll replace sin x by another variable t:

2t^2 - t - 1 = 0

We'll solve the quadratic:

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

t1 = (1+3)/4

t1 = 1

t2 = -1/2

We'll put sin x = t1 => sin x = 1

x = k*pi/2 + k*pi

sin x = t2

sin x = -1/2

x = (-1)^(k+1)*pi/6 + k*pi

**The solutions of the equation are: {k*pi/2 + k*pi} U {(-1)^(k+1)*pi/6 + k*pi}.**