What is the general solution for sin^2(x)=((10m-5))/2 ?

The general solution for sin^2(x)=((10m-5))/2 is

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sin^2(x)=((10m-5))/2

sin^{2} (x) = \frac{( 10 m - 5 )}{2}

x = arcsin (\frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (\frac{10 m - 5}{2}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 10 m - 5 )}{2}

Solve by substitution

sin^{2} (x) = \frac{10 m - 5}{2}

Let:

sin (x) = u

u^{2} = \frac{10 m - 5}{2}

u^{2} = \frac{10 m - 5}{2} : u = \frac{10 m - 5}{2}, u = - \frac{10 m - 5}{2}

u^{2} = \frac{10 m - 5}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{10 m - 5}{2}, u = - \frac{10 m - 5}{2}

Substitute back

u = sin (x)

sin (x) = \frac{10 m - 5}{2}, sin (x) = - \frac{10 m - 5}{2}

sin (x) = \frac{10 m - 5}{2}, sin (x) = - \frac{10 m - 5}{2}

sin (x) = \frac{10 m - 5}{2} : x = arcsin (\frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (- \frac{10 m - 5}{2}) + 2 πn

sin (x) = \frac{10 m - 5}{2}

Apply trig inverse properties

sin (x) = \frac{10 m - 5}{2}

General solutions for

sin (x) = \frac{10 m - 5}{2}

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

x = arcsin (\frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (- \frac{10 m - 5}{2}) + 2 πn

x = arcsin (\frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (- \frac{10 m - 5}{2}) + 2 πn

sin (x) = - \frac{10 m - 5}{2} : x = arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (\frac{10 m - 5}{2}) + 2 πn

sin (x) = - \frac{10 m - 5}{2}

Apply trig inverse properties

sin (x) = - \frac{10 m - 5}{2}

General solutions for

sin (x) = - \frac{10 m - 5}{2}

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

x = arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (\frac{10 m - 5}{2}) + 2 πn

x = arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (\frac{10 m - 5}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = arcsin (- \frac{10 m - 5}{2}) + 2 πn, x = π + arcsin (\frac{10 m - 5}{2}) + 2 πn

cos (3 x) + sin (x) = 0

7 tan (x) = 23 + tan (x)

sin (θ) - 3 cos (θ) = 0

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

sin (x) cos (x) cos (2 x) = \frac{1}{8}

What is the general solution for sin^2(x)=((10m-5))/2 ?
The general solution for sin^2(x)=((10m-5))/2 is