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

The general solution for -tan(2θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(4pin)/3 ,θ=(2pi)/3+(4pin)/3

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-tan(2θ)+2sin(θ)=0

- tan (2 θ) + 2 sin (θ) = 0

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

Solution steps

- tan (2 θ) + 2 sin (θ) = 0

Express with sin, cos

\frac{- sin ( 2 θ ) + 2 cos ( 2 θ ) sin ( θ )}{cos ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin (2 θ) + 2 cos (2 θ) sin (θ) = 0

Rewrite using trig identities

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) = 0

Factor

2 cos (2 θ) sin (θ) - 2 cos (θ) sin (θ) : 2 sin (θ) (cos (2 θ) - cos (θ))

2 sin (θ) (cos (2 θ) - cos (θ)) = 0

Solving each part separately

sin (θ) = 0 or cos (2 θ) - cos (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

cos (2 θ) - cos (θ) = 0 : θ = \frac{4 πn}{3}, θ = \frac{2 π}{3} + \frac{4 πn}{3}

Combine all the solutions

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

- 3 cos^{2} (θ) + 6 = sin (θ) + 5

cos^{2} (θ) + 4 cos (θ) = 0

2 sin (x) = 2

tan (θ) = \frac{- 1}{3}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

What is the general solution for -tan(2θ)+2sin(θ)=0 ?
The general solution for -tan(2θ)+2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=(4pin)/3 ,θ=(2pi)/3+(4pin)/3