What is the general solution for (1+cos(x))(1-sin(x))-cos^2(x)=0 ?

The general solution for (1+cos(x))(1-sin(x))-cos^2(x)=0 is x= pi/2+2pin,x= pi/4+pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{π}{4} + πn

Solution steps

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{π}{4} + πn

cos (x) = - 8

arctan (\frac{18.5}{x}) - arctan (\frac{42.41}{x}) = 21.27

2 sin (2 x) = sin^{2} (x)

5 sin^{2} (x) = 3 sin (x) + 2

sin^{2} (x) = - (\frac{1}{2})

What is the general solution for (1+cos(x))(1-sin(x))-cos^2(x)=0 ?
The general solution for (1+cos(x))(1-sin(x))-cos^2(x)=0 is x= pi/2+2pin,x= pi/4+pin