What is the general solution for sin^2(x)=((10m-7))/9 ?

The general solution for sin^2(x)=((10m-7))/9 is

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sin^2(x)=((10m-7))/9

sin^{2} (x) = \frac{( 10 m - 7 )}{9}

x = arcsin (\frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (\frac{10 m - 7}{3}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 10 m - 7 )}{9}

Solve by substitution

sin^{2} (x) = \frac{10 m - 7}{9}

Let:

sin (x) = u

u^{2} = \frac{10 m - 7}{9}

u^{2} = \frac{10 m - 7}{9} : u = \frac{10 m - 7}{3}, u = - \frac{10 m - 7}{3}

u^{2} = \frac{10 m - 7}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{10 m - 7}{9}, u = - \frac{10 m - 7}{9}

Simplify

\frac{10 m - 7}{9} : \frac{10 m - 7}{3}

\frac{10 m - 7}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{10 m - 7}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{10 m - 7}{3}

Simplify

- \frac{10 m - 7}{9} : - \frac{10 m - 7}{3}

- \frac{10 m - 7}{9}

Simplify

\frac{10 m - 7}{9} : \frac{10 m - 7}{3}

\frac{10 m - 7}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{10 m - 7}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{10 m - 7}{3}

= - \frac{10 m - 7}{3}

u = \frac{10 m - 7}{3}, u = - \frac{10 m - 7}{3}

Substitute back

u = sin (x)

sin (x) = \frac{10 m - 7}{3}, sin (x) = - \frac{10 m - 7}{3}

sin (x) = \frac{10 m - 7}{3}, sin (x) = - \frac{10 m - 7}{3}

sin (x) = \frac{10 m - 7}{3} : x = arcsin (\frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (- \frac{10 m - 7}{3}) + 2 πn

sin (x) = \frac{10 m - 7}{3}

Apply trig inverse properties

sin (x) = \frac{10 m - 7}{3}

General solutions for

sin (x) = \frac{10 m - 7}{3}

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

x = arcsin (\frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (- \frac{10 m - 7}{3}) + 2 πn

x = arcsin (\frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (- \frac{10 m - 7}{3}) + 2 πn

sin (x) = - \frac{10 m - 7}{3} : x = arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (\frac{10 m - 7}{3}) + 2 πn

sin (x) = - \frac{10 m - 7}{3}

Apply trig inverse properties

sin (x) = - \frac{10 m - 7}{3}

General solutions for

sin (x) = - \frac{10 m - 7}{3}

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

x = arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (\frac{10 m - 7}{3}) + 2 πn

x = arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (\frac{10 m - 7}{3}) + 2 πn

Combine all the solutions

x = arcsin (\frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = arcsin (- \frac{10 m - 7}{3}) + 2 πn, x = π + arcsin (\frac{10 m - 7}{3}) + 2 πn

t, f = cos^{2} (a t)

8 sin (x) = 2 + \frac{4}{csc ( x )}

y, 2 e^{x} - sin (y) = x

sec^{2} (x) = \frac{1}{2}

3 sin^{2} (x) = 3 cos^{2} (x)

What is the general solution for sin^2(x)=((10m-7))/9 ?
The general solution for sin^2(x)=((10m-7))/9 is