What is the general solution for sin^2(x)=((3m-2))/8 ?

The general solution for sin^2(x)=((3m-2))/8 is

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sin^2(x)=((3m-2))/8

sin^{2} (x) = \frac{( 3 m - 2 )}{8}

x = arcsin (\frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (- \frac{2 3 m - 2}{4}) + 2 πn, x = arcsin (- \frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (\frac{2 3 m - 2}{4}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 3 m - 2 )}{8}

Solve by substitution

sin (x) = \frac{2 3 m - 2}{4}, sin (x) = - \frac{2 3 m - 2}{4}

sin (x) = \frac{2 3 m - 2}{4} : x = arcsin (\frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (- \frac{2 3 m - 2}{4}) + 2 πn

sin (x) = - \frac{2 3 m - 2}{4} : x = arcsin (- \frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (\frac{2 3 m - 2}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (- \frac{2 3 m - 2}{4}) + 2 πn, x = arcsin (- \frac{2 3 m - 2}{4}) + 2 πn, x = π + arcsin (\frac{2 3 m - 2}{4}) + 2 πn

arccos (x) = arcsin (\frac{1}{3}) + arccos (\frac{1}{4})

sin (2 x) = cos (4 0^{\circ})

- 15 sin (x) - 8 cos (x) = 10

- 5 sec^{2} (x) + 20 = 0

1 + cos (x) = 3 \cdot sin (x)

What is the general solution for sin^2(x)=((3m-2))/8 ?
The general solution for sin^2(x)=((3m-2))/8 is