What is the general solution for sin(2θ)-cos(2θ)=0 ?

The general solution for sin(2θ)-cos(2θ)=0 is θ= pi/8+(pin)/2

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sin(2θ)-cos(2θ)=0

sin (2 θ) - cos (2 θ) = 0

θ = \frac{π}{8} + \frac{πn}{2}

Solution steps

sin (2 θ) - cos (2 θ) = 0

Rewrite using trig identities

tan (2 θ) - 1 = 0

Move

1

to the right side

tan (2 θ) = 1

General solutions for

tan (2 θ) = 1

2 θ = \frac{π}{4} + πn

Solve

2 θ = \frac{π}{4} + πn : θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

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

4 csc (θ) - 5 = 0

tan (4 x) + 3 = 0

2 sin (x) + 4 = 5

x, z = y sin (4 x)

What is the general solution for sin(2θ)-cos(2θ)=0 ?
The general solution for sin(2θ)-cos(2θ)=0 is θ= pi/8+(pin)/2