What is the general solution for sin(2θ)+cos(2θ)=1 ?

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

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sin(2θ)+cos(2θ)=1

sin (2 θ) + cos (2 θ) = 1

θ = πn, θ = πn + \frac{π}{4}

Solution steps

sin (2 θ) + cos (2 θ) = 1

Rewrite using trig identities

2 sin (2 θ + \frac{π}{4}) = 1

Divide both sides by

2

sin (2 θ + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (2 θ + \frac{π}{4}) = \frac{2}{2}

2 θ + \frac{π}{4} = \frac{π}{4} + 2 πn, 2 θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

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

Solve

2 θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn : θ = πn + \frac{π}{4}

θ = πn, θ = πn + \frac{π}{4}

tan (4 x) = 3, 0 \leq x \leq 2 π

6 arctan (x) = 2 π

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

7 csc (x) - 3 = 11

2 cos^{3} (x) - 1 = 0

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