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

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

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

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

x = \frac{3 π}{8} + \frac{πn}{2}

Solution steps

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

Rewrite using trig identities

1 + tan (2 x) = 0

Move

1

to the right side

tan (2 x) = - 1

General solutions for

tan (2 x) = - 1

2 x = \frac{3 π}{4} + πn

Solve

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

x = \frac{3 π}{8} + \frac{πn}{2}

sin (x) = \frac{4}{7}

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

4 sin (θ) + 5 = 3

(2 cos (x) + 3) (2 sin (x) + 1) = 0

sin^{2} (x) = 2 sin (x) + 3, 0 \leq x \leq 2 π

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