What is the general solution for -7sin^2(θ)+2sin(θ)+7=0 ?

The general solution for -7sin^2(θ)+2sin(θ)+7=0 is θ=-1.04974…+2pin,θ=pi+1.04974…+2pin

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-7sin^2(θ)+2sin(θ)+7=0

- 7 sin^{2} (θ) + 2 sin (θ) + 7 = 0

θ = - 1.04974 \dots + 2 πn, θ = π + 1.04974 \dots + 2 πn

Solution steps

- 7 sin^{2} (θ) + 2 sin (θ) + 7 = 0

Solve by substitution

sin (θ) = - \frac{- 1 + 5 2}{7}, sin (θ) = \frac{1 + 5 2}{7}

sin (θ) = - \frac{- 1 + 5 2}{7} : θ = arcsin (- \frac{- 1 + 5 2}{7}) + 2 πn, θ = π + arcsin (\frac{- 1 + 5 2}{7}) + 2 πn

sin (θ) = \frac{1 + 5 2}{7} :

No Solution

Combine all the solutions

θ = arcsin (- \frac{- 1 + 5 2}{7}) + 2 πn, θ = π + arcsin (\frac{- 1 + 5 2}{7}) + 2 πn

Show solutions in decimal form

θ = - 1.04974 \dots + 2 πn, θ = π + 1.04974 \dots + 2 πn

cos (x) - 2 sin^{2} (x) + 1 = 0

3 tan (2 x) + 3 = 0, 0 \leq x < 2 π

sin (2 x) - sin (x) cos (x) = cos (x)

cot^{2} (x) = 1 - csc (x)

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

What is the general solution for -7sin^2(θ)+2sin(θ)+7=0 ?
The general solution for -7sin^2(θ)+2sin(θ)+7=0 is θ=-1.04974…+2pin,θ=pi+1.04974…+2pin