# solve the equation 2-2sinx = cos^2 x when x is between -180 and 180 degrees thanks

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You need to write the equation in terms of sin x, hence, you should use the fundamental formula of trigonometry such that:

`2 - 2sinx = cos^2 x => 2 - 2sinx = 1 - sin^2 x`

You need to factor out 2 to the left side such that:

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

You should write the difference of squares to the right as a product such that:

`2(1 - sin x) = (1 - sin x)(1 + sin x)`

You need to move all terms to one side such that:

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

You need to factor out `(1 - sin x) ` such that:

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

`1 - sin x = 0 => -sin x = -1 => sin x = 1 => x = pi/2 in [-pi,pi]`

**Hence, evaluating the solution to the given equation, under the given conditions, yields `x = pi/2` .**

2-2sinx=cos^2x

2-2sinx=1-sin^2x

By rearranging;

sin^2x-2sinx+1=0

(sinx-1)^2=0 Factorised;

Therefore; Sinx=1

The minor solution=a=pi/2

General solution of the equation is given by;

x=n*pi+(-1)^n*a

x=n*pi+(-1)^n*pi/2 (n is an integer)

When n=0;

x=pi/2 satisfied

When n=1;

x=pi/2 satisfied

When n=2;

x=5pi/2 unsatisfied

When n=-1;

x=-3pi/2 unsatisfied

**Therefore the solutions of the given equation in the range of -180 to 180 = pi/2=90**