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

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

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

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

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

Solution steps

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

Subtract

sin (2 x)

from both sides

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

Combine all the solutions

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

cos (θ) = 5

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

x, tan (3 x) = 5 tan (x)

3 sin (2 t) = 0

tan (θ) = \frac{8}{5}

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