What is the general solution for 2sin^2(β)=9sin(β)-4 ?

The general solution for 2sin^2(β)=9sin(β)-4 is β= pi/6+2pin,β=(5pi)/6+2pin

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2sin^2(β)=9sin(β)-4

2 sin^{2} (β) = 9 sin (β) - 4

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

Solution steps

2 sin^{2} (β) = 9 sin (β) - 4

Solve by substitution

sin (β) = 4, sin (β) = \frac{1}{2}

sin (β) = 4 :

No Solution

sin (β) = \frac{1}{2} : β = \frac{π}{6} + 2 πn, β = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

4 cos^{2} (x) - 1 = 4, cos (2 x) - 1 = 0

sin (2 θ) = 0.4

0.5 = sin (x) cos (x)

sin (4 x) - cos (4 x) = \frac{1}{2}

tan^{2} (θ) + sec (θ) = - 2

What is the general solution for 2sin^2(β)=9sin(β)-4 ?
The general solution for 2sin^2(β)=9sin(β)-4 is β= pi/6+2pin,β=(5pi)/6+2pin