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

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

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-tan(2θ)+2sin(θ)=0,0<= θ<2pi

- tan (2 θ) + 2 sin (θ) = 0, 0 \leq θ < 2 π

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Solution steps

- tan (2 θ) + 2 sin (θ) = 0, 0 \leq θ < 2 π

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, 0 \leq θ < 2 π : θ = 0, θ = π

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

Combine all the solutions

θ = 0, θ = π, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

cos (x) = \frac{2 tan ( x )}{1 + tan ^{2} ( x )}

(1 + cos (4 x)) sin (2 x) = cos^{2} (2 x)

cos (2 x) - cos (6 x) - sin (4 x) = 0

cot (θ) = \frac{15}{8}

2 sin (2 x) = sin (4 x)

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