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

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

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2sin(θ)+1=csc(θ)

2 sin (θ) + 1 = csc (θ)

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

Solution steps

2 sin (θ) + 1 = csc (θ)

Subtract

csc (θ)

from both sides

2 sin (θ) + 1 - 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

cot (x) - 1 = csc (x)

- 6 cos^{2} (θ) - 3 cos (θ) + 1 = - 4 cos (θ)

tan (θ) = 1.2

sin (θ) = - \frac{15}{17}

tan (θ) = - \frac{3}{2}

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