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

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

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

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

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

u^{2} = \frac{9 m - 8}{2}

u^{2} = \frac{9 m - 8}{2} : u = \frac{9 m - 8}{2}, u = - \frac{9 m - 8}{2}

u^{2} = \frac{9 m - 8}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{9 m - 8}{2}, u = - \frac{9 m - 8}{2}

Substitute back

u = sin (x)

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

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

sin (x) = \frac{9 m - 8}{2} : x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn

sin (x) = - \frac{9 m - 8}{2} : x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = arcsin (- \frac{9 m - 8}{2}) + 2 πn, x = π + arcsin (\frac{9 m - 8}{2}) + 2 πn

\frac{5}{2} cos (2 x) - \frac{1}{2} = 3 sin (x)

cos (π + x) = \frac{1}{2}

(tan (x) - 4) (16 sin^{2} (x) - 1) = 0

sec^{2} (x) = 2 sin^{2} (x) + tan^{2} (x)

64 cos^{2} (θ) = 49

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