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

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

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2cos(θ)+sin(2θ)=0

2 cos (θ) + sin (2 θ) = 0

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

Solution steps

2 cos (θ) + sin (2 θ) = 0

Rewrite using trig identities

2 cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

2 cos (θ) + 2 cos (θ) sin (θ) : 2 cos (θ) (sin (θ) + 1)

2 cos (θ) (sin (θ) + 1) = 0

Solving each part separately

cos (θ) = 0 or sin (θ) + 1 = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

tan (x) = \frac{6 \cdot 3}{6}

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

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

2 + cos (2 x) = - 3 cos (x)

tan (\frac{3 x}{2} + \frac{π}{2}) = 1

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