What is the general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 ?

The general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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2sin(2θ)-3-6/(sin(2θ)-1)=0

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

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

Solution steps

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0

Solve by substitution

sin (2 θ) = 3, sin (2 θ) = - \frac{1}{2}

sin (2 θ) = 3 :

No Solution

sin (2 θ) = - \frac{1}{2} : θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Combine all the solutions

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

8 cos (B) + 52 = 8

sin (a x) = 0

cos (x + 3 0^{\circ}) = 2 cos (x)

x, sec (x) tan (x) = 23

4 sin (θ) - 4 cos (θ) = 2

What is the general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 ?
The general solution for 2sin(2θ)-3-6/(sin(2θ)-1)=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin