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{π}{2} + 2 πn

Combine all the solutions

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

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

2 cos (2 x) + 1 + 4 sin (x) = 0

tan (x) = - 1, tan (π + x) = y

sin (θ) sec (θ) - sin (θ) = 0

r = 2 a cos (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