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

The general solution for sin(2θ)+cos(2θ)=0 is θ=(3pi)/8+(pin)/2

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sin(2θ)+cos(2θ)=0

sin (2 θ) + cos (2 θ) = 0

θ = \frac{3 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{3 π}{4} + πn

Solve

2 θ = \frac{3 π}{4} + πn : θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

5 cot^{2} (x) - 15 = 0

sin (x) = \frac{2}{5}

csc (x) = 5

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

tan^{2} (x) - 3 tan (x) = 0, 0 \leq x < 2 π

What is the general solution for sin(2θ)+cos(2θ)=0 ?
The general solution for sin(2θ)+cos(2θ)=0 is θ=(3pi)/8+(pin)/2