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

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

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

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

1 - cos (2 x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

sin (θ) = \frac{5}{6}, cos (θ)

tan^{2} (t) - sec^{2} (t) = 1

sin (x) tan (x) = \frac{1}{cot ( x )}

0 = - 2 sin (x) - 1

x, z = sin (2 x + 3 y)

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