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

The general solution for sin(θ)+cos(2θ)=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin

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sin(θ)+cos(2θ)=0

sin (θ) + cos (2 θ) = 0

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

Solution steps

sin (θ) + cos (2 θ) = 0

Rewrite using trig identities

1 + sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 1

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

3 sec (θ) - 4 = 0

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

cot (2 θ) = \frac{3}{4}

x, 0 = 4 - sec^{2} (x)

tan (x) = - \frac{4}{33}

What is the general solution for sin(θ)+cos(2θ)=0 ?
The general solution for sin(θ)+cos(2θ)=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin