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

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

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

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

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

Solution steps

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

Solving each part separately

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

1 - sin (2 x) = 0 : x = \frac{π}{4} + πn

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

Combine all the solutions

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

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∣ (211) ∣ ∣ (- 37 - 1) ∣ (cos (x)) = 0

u, x = 4 sinh (u)

tan (x) = \frac{25}{32}

sin (x + 2 0^{\circ}) + sin (4 0^{\circ} - x) = 1

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