What is the general solution for sin^2(x)=((7m-9))/9 ?

The general solution for sin^2(x)=((7m-9))/9 is

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sin^2(x)=((7m-9))/9

sin^{2} (x) = \frac{( 7 m - 9 )}{9}

x = arcsin (\frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (\frac{7 m - 9}{3}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 7 m - 9 )}{9}

Solve by substitution

sin^{2} (x) = \frac{7 m - 9}{9}

Let:

sin (x) = u

u^{2} = \frac{7 m - 9}{9}

u^{2} = \frac{7 m - 9}{9} : u = \frac{7 m - 9}{3}, u = - \frac{7 m - 9}{3}

u^{2} = \frac{7 m - 9}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{7 m - 9}{9}, u = - \frac{7 m - 9}{9}

Simplify

\frac{7 m - 9}{9} : \frac{7 m - 9}{3}

\frac{7 m - 9}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7 m - 9}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{7 m - 9}{3}

Simplify

- \frac{7 m - 9}{9} : - \frac{7 m - 9}{3}

- \frac{7 m - 9}{9}

Simplify

\frac{7 m - 9}{9} : \frac{7 m - 9}{3}

\frac{7 m - 9}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7 m - 9}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{7 m - 9}{3}

= - \frac{7 m - 9}{3}

u = \frac{7 m - 9}{3}, u = - \frac{7 m - 9}{3}

Substitute back

u = sin (x)

sin (x) = \frac{7 m - 9}{3}, sin (x) = - \frac{7 m - 9}{3}

sin (x) = \frac{7 m - 9}{3}, sin (x) = - \frac{7 m - 9}{3}

sin (x) = \frac{7 m - 9}{3} : x = arcsin (\frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (- \frac{7 m - 9}{3}) + 2 πn

sin (x) = \frac{7 m - 9}{3}

Apply trig inverse properties

sin (x) = \frac{7 m - 9}{3}

General solutions for

sin (x) = \frac{7 m - 9}{3}

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

x = arcsin (\frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (- \frac{7 m - 9}{3}) + 2 πn

x = arcsin (\frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (- \frac{7 m - 9}{3}) + 2 πn

sin (x) = - \frac{7 m - 9}{3} : x = arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (\frac{7 m - 9}{3}) + 2 πn

sin (x) = - \frac{7 m - 9}{3}

Apply trig inverse properties

sin (x) = - \frac{7 m - 9}{3}

General solutions for

sin (x) = - \frac{7 m - 9}{3}

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

x = arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (\frac{7 m - 9}{3}) + 2 πn

x = arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (\frac{7 m - 9}{3}) + 2 πn

Combine all the solutions

x = arcsin (\frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = arcsin (- \frac{7 m - 9}{3}) + 2 πn, x = π + arcsin (\frac{7 m - 9}{3}) + 2 πn

sin (x) = 2, 0 < x < 2 π

2 csc^{2} (x) - 2 csc (x) - 1 = 0

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

cos (x) = \frac{6}{15}

arccos (x) = 90

What is the general solution for sin^2(x)=((7m-9))/9 ?
The general solution for sin^2(x)=((7m-9))/9 is