What is the general solution for (sin(x)-cos(x))^2=1^2 ?

The general solution for (sin(x)-cos(x))^2=1^2 is x=2pin+(3pi)/2 ,x=2pin+2pi,x= pi/2+2pin,x=2pin+pi

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

(sin (x) - cos (x))^{2} = 1^{2}

x = 2 πn + \frac{3 π}{2}, x = 2 πn + 2 π, x = \frac{π}{2} + 2 πn, x = 2 πn + π

Solution steps

(sin (x) - cos (x))^{2} = 1^{2}

Subtract

1^{2}

from both sides

(sin (x) - cos (x))^{2} - 1 = 0

Factor

(sin (x) - cos (x))^{2} - 1 : (sin (x) - cos (x) + 1) (sin (x) - cos (x) - 1)

(sin (x) - cos (x) + 1) (sin (x) - cos (x) - 1) = 0

Solving each part separately

sin (x) - cos (x) + 1 = 0 or sin (x) - cos (x) - 1 = 0

sin (x) - cos (x) + 1 = 0 : x = 2 πn + \frac{3 π}{2}, x = 2 πn + 2 π

sin (x) - cos (x) - 1 = 0 : x = \frac{π}{2} + 2 πn, x = 2 πn + π

Combine all the solutions

x = 2 πn + \frac{3 π}{2}, x = 2 πn + 2 π, x = \frac{π}{2} + 2 πn, x = 2 πn + π

tan (\frac{π}{2} - x) + 2 tan (\frac{π}{2} + x) = 1

\frac{11.0994}{cos ( x )} = \frac{3.5355}{sin ( x )}

(\frac{sec ( x )}{cos ( x )})^{2} = tan^{2} (x)

sin (\frac{x}{2}) - cos (\frac{x}{2}) = 0

2 sin^{2} (\frac{x}{3}) = 1, - π \leq, x \leq π

What is the general solution for (sin(x)-cos(x))^2=1^2 ?
The general solution for (sin(x)-cos(x))^2=1^2 is x=2pin+(3pi)/2 ,x=2pin+2pi,x= pi/2+2pin,x=2pin+pi