What is the general solution for sin^2(x)=((5m-3))/3 ?

The general solution for sin^2(x)=((5m-3))/3 is

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sin^2(x)=((5m-3))/3

sin^{2} (x) = \frac{( 5 m - 3 )}{3}

x = arcsin (\frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (\frac{5 m - 3}{3}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 5 m - 3 )}{3}

Solve by substitution

sin^{2} (x) = \frac{5 m - 3}{3}

Let:

sin (x) = u

u^{2} = \frac{5 m - 3}{3}

u^{2} = \frac{5 m - 3}{3} : u = \frac{5 m - 3}{3}, u = - \frac{5 m - 3}{3}

u^{2} = \frac{5 m - 3}{3}

For

x^{2} = f (a)

the solutions are

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

u = \frac{5 m - 3}{3}, u = - \frac{5 m - 3}{3}

Substitute back

u = sin (x)

sin (x) = \frac{5 m - 3}{3}, sin (x) = - \frac{5 m - 3}{3}

sin (x) = \frac{5 m - 3}{3}, sin (x) = - \frac{5 m - 3}{3}

sin (x) = \frac{5 m - 3}{3} : x = arcsin (\frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (- \frac{5 m - 3}{3}) + 2 πn

sin (x) = \frac{5 m - 3}{3}

Apply trig inverse properties

sin (x) = \frac{5 m - 3}{3}

General solutions for

sin (x) = \frac{5 m - 3}{3}

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

x = arcsin (\frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (- \frac{5 m - 3}{3}) + 2 πn

x = arcsin (\frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (- \frac{5 m - 3}{3}) + 2 πn

sin (x) = - \frac{5 m - 3}{3} : x = arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (\frac{5 m - 3}{3}) + 2 πn

sin (x) = - \frac{5 m - 3}{3}

Apply trig inverse properties

sin (x) = - \frac{5 m - 3}{3}

General solutions for

sin (x) = - \frac{5 m - 3}{3}

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

x = arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (\frac{5 m - 3}{3}) + 2 πn

x = arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (\frac{5 m - 3}{3}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = arcsin (- \frac{5 m - 3}{3}) + 2 πn, x = π + arcsin (\frac{5 m - 3}{3}) + 2 πn

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

cos (x) = \frac{4}{3.3 8 ^{2} + 2 ^{2}}

- 3 = 6 sin (θ)

cos (x) = \frac{- 4}{5}

\frac{1}{cos ( θ )} + tan (θ) = 2

What is the general solution for sin^2(x)=((5m-3))/3 ?
The general solution for sin^2(x)=((5m-3))/3 is