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

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

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

solve for

x, 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

cos (θ) = \frac{12}{20}

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

cos (θ) = \frac{12}{15}

arctan (x) = 45

tan^{2} (x) + cot^{2} (x) - 2 = 0

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