What is the general solution for 1+2sin(θ)=csc(θ) ?

The general solution for 1+2sin(θ)=csc(θ) is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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1+2sin(θ)=csc(θ)

1 + 2 sin (θ) = csc (θ)

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

1 + 2 sin (θ) = csc (θ)

Subtract

csc (θ)

from both sides

1 + 2 sin (θ) - csc (θ) = 0

Rewrite using trig identities

1 - csc (θ) + \frac{2}{csc ( θ )} = 0

Solve by substitution

csc (θ) = - 1, csc (θ) = 2

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = 2 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

- 6 = - 4 + sec (x + \frac{π}{4})

2 sin^{2} (x) + 3 sin (x) = 2

tan (x) sec (x) + 3 tan (x) = 0

\frac{csc ( b )}{csc ( b )} = \frac{1}{1 - sin ^{2} ( b )}

3 sec (4 x) = 2

What is the general solution for 1+2sin(θ)=csc(θ) ?
The general solution for 1+2sin(θ)=csc(θ) is θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin