What is the general solution for sin^2(x)=((5m-7))/6 ?

The general solution for sin^2(x)=((5m-7))/6 is

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sin^2(x)=((5m-7))/6

sin^{2} (x) = \frac{( 5 m - 7 )}{6}

x = arcsin (\frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (\frac{5 m - 7}{6}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 5 m - 7 )}{6}

Solve by substitution

sin^{2} (x) = \frac{5 m - 7}{6}

Let:

sin (x) = u

u^{2} = \frac{5 m - 7}{6}

u^{2} = \frac{5 m - 7}{6} : u = \frac{5 m - 7}{6}, u = - \frac{5 m - 7}{6}

u^{2} = \frac{5 m - 7}{6}

For

x^{2} = f (a)

the solutions are

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

u = \frac{5 m - 7}{6}, u = - \frac{5 m - 7}{6}

Substitute back

u = sin (x)

sin (x) = \frac{5 m - 7}{6}, sin (x) = - \frac{5 m - 7}{6}

sin (x) = \frac{5 m - 7}{6}, sin (x) = - \frac{5 m - 7}{6}

sin (x) = \frac{5 m - 7}{6} : x = arcsin (\frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (- \frac{5 m - 7}{6}) + 2 πn

sin (x) = \frac{5 m - 7}{6}

Apply trig inverse properties

sin (x) = \frac{5 m - 7}{6}

General solutions for

sin (x) = \frac{5 m - 7}{6}

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

x = arcsin (\frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (- \frac{5 m - 7}{6}) + 2 πn

x = arcsin (\frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (- \frac{5 m - 7}{6}) + 2 πn

sin (x) = - \frac{5 m - 7}{6} : x = arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (\frac{5 m - 7}{6}) + 2 πn

sin (x) = - \frac{5 m - 7}{6}

Apply trig inverse properties

sin (x) = - \frac{5 m - 7}{6}

General solutions for

sin (x) = - \frac{5 m - 7}{6}

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

x = arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (\frac{5 m - 7}{6}) + 2 πn

x = arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (\frac{5 m - 7}{6}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = arcsin (- \frac{5 m - 7}{6}) + 2 πn, x = π + arcsin (\frac{5 m - 7}{6}) + 2 πn

sin (θ) = \frac{- 4}{4 2}, csc (θ) < 0

sin^{2} (x) = 0.2

sin (x) = \frac{sin ( 4 5 ^{\circ} )}{1.3}

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

(2 cos (x) + 1) (sin (x) - 1) = 0

What is the general solution for sin^2(x)=((5m-7))/6 ?
The general solution for sin^2(x)=((5m-7))/6 is