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

The general solution for 2sin(θ)-tan^2(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.89590…+2pin,θ=pi-0.89590…+2pin

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2sin(θ)-tan^2(θ)=0

2 sin (θ) - tan^{2} (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

Solution steps

2 sin (θ) - tan^{2} (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

Factor

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ) : sin (θ) (- sin (θ) + 2 cos^{2} (θ))

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

Solving each part separately

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

- sin (θ) + 2 cos^{2} (θ) = 0 : θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

sin (θ) - 3 cos (θ) = 1

cos (a) = \frac{1}{2}

cos^{2} (θ) = \frac{3}{4}

tan (θ) = \frac{3}{8}

9 cos (x) + 5 = - cos (x)

What is the general solution for 2sin(θ)-tan^2(θ)=0 ?
The general solution for 2sin(θ)-tan^2(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.89590…+2pin,θ=pi-0.89590…+2pin