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

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

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sin(2θ)+2cos(θ)=0

sin (2 θ) + 2 cos (θ) = 0

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

Solution steps

sin (2 θ) + 2 cos (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

cot (3 x) = 0, 0 \leq x \leq 2 π

sec (2 x) = 2, 0 \leq x \leq 2 π

sin (x - \frac{π}{4}) = 0

cot (x - \frac{π}{2}) = 1

sec (x) = tan (x)

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