What is the general solution for -9cos^2(θ)+1=sin(θ) ?

The general solution for -9cos^2(θ)+1=sin(θ) is θ= pi/2+2pin,θ=-1.09491…+2pin,θ=pi+1.09491…+2pin

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-9cos^2(θ)+1=sin(θ)

- 9 cos^{2} (θ) + 1 = sin (θ)

θ = \frac{π}{2} + 2 πn, θ = - 1.09491 \dots + 2 πn, θ = π + 1.09491 \dots + 2 πn

Solution steps

- 9 cos^{2} (θ) + 1 = sin (θ)

Subtract

sin (θ)

from both sides

- 9 cos^{2} (θ) + 1 - sin (θ) = 0

Rewrite using trig identities

- 8 - sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{8}{9}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - \frac{8}{9} : θ = arcsin (- \frac{8}{9}) + 2 πn, θ = π + arcsin (\frac{8}{9}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = arcsin (- \frac{8}{9}) + 2 πn, θ = π + arcsin (\frac{8}{9}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = - 1.09491 \dots + 2 πn, θ = π + 1.09491 \dots + 2 πn

1 - sin (θ) = sin (θ)

cot (\frac{θ}{3} + \frac{3 π}{4}) = 3

cos (x) = \frac{tan ( x ) + cot ( x )}{csc ( x )}

- tan^{2} (x) - 1 = 0

2 cot^{2} (y) + 3 csc (y) = 0

What is the general solution for -9cos^2(θ)+1=sin(θ) ?
The general solution for -9cos^2(θ)+1=sin(θ) is θ= pi/2+2pin,θ=-1.09491…+2pin,θ=pi+1.09491…+2pin