What is the general solution for sin^2(x)=((4m-9))/8 ?

The general solution for sin^2(x)=((4m-9))/8 is

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sin^2(x)=((4m-9))/8

sin^{2} (x) = \frac{( 4 m - 9 )}{8}

x = arcsin (\frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (- \frac{2 4 m - 9}{4}) + 2 πn, x = arcsin (- \frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (\frac{2 4 m - 9}{4}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 4 m - 9 )}{8}

Solve by substitution

sin (x) = \frac{2 4 m - 9}{4}, sin (x) = - \frac{2 4 m - 9}{4}

sin (x) = \frac{2 4 m - 9}{4} : x = arcsin (\frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (- \frac{2 4 m - 9}{4}) + 2 πn

sin (x) = - \frac{2 4 m - 9}{4} : x = arcsin (- \frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (\frac{2 4 m - 9}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (- \frac{2 4 m - 9}{4}) + 2 πn, x = arcsin (- \frac{2 4 m - 9}{4}) + 2 πn, x = π + arcsin (\frac{2 4 m - 9}{4}) + 2 πn

- csc (x) = 1

sin (θ) = - \frac{7}{4}

x, sin (x) = e^{y}

sin^{4} (x) - sin^{2} (x) + 1 = 0

tan^{3} (x) - tan^{2} (x) - 3 tan (x) + 3 = 0, tan (x) > 1

What is the general solution for sin^2(x)=((4m-9))/8 ?
The general solution for sin^2(x)=((4m-9))/8 is