What is the general solution for 4csc(2θ)+8=0 ?

The general solution for 4csc(2θ)+8=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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4csc(2θ)+8=0

4 csc (2 θ) + 8 = 0

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Solution steps

4 csc (2 θ) + 8 = 0

Move

8

to the right side

4 csc (2 θ) = - 8

Divide both sides by

4

csc (2 θ) = - 2

General solutions for

csc (2 θ) = - 2

2 θ = \frac{7 π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

tan (x) = - \frac{4}{7}

2 sin (x) + 3 cos (x) = 3

cos (θ) = \frac{π}{3}

2 cos^{2} (x) = \frac{1}{2}

2 sin (2 x + \frac{3 π}{2}) = 3

What is the general solution for 4csc(2θ)+8=0 ?
The general solution for 4csc(2θ)+8=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin