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

The general solution for cos(2θ)+sin(2θ)=-1 is θ=pin+pi/2 ,θ=pin+(3pi)/4

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cos(2θ)+sin(2θ)=-1

cos (2 θ) + sin (2 θ) = - 1

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

Solution steps

cos (2 θ) + sin (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{5 π}{4} + 2 πn, 2 θ + \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

2 θ + \frac{π}{4} = \frac{5 π}{4} + 2 πn : θ = πn + \frac{π}{2}

Solve

2 θ + \frac{π}{4} = \frac{7 π}{4} + 2 πn : θ = πn + \frac{3 π}{4}

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

sin (θ) = \frac{19}{27}

sin^{4} (x) = \frac{1}{2} cos^{2} (x)

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

sinh^{2} (x) + 3 tanh^{2} (x) = 4

\frac{sin ( 5 1 ^{\circ} )}{18} = \frac{sin ( c )}{17}

What is the general solution for cos(2θ)+sin(2θ)=-1 ?
The general solution for cos(2θ)+sin(2θ)=-1 is θ=pin+pi/2 ,θ=pin+(3pi)/4