What is the general solution for sin^2(x)=((8m-2))/9 ?

The general solution for sin^2(x)=((8m-2))/9 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(x)=((8m-2))/9

sin^{2} (x) = \frac{( 8 m - 2 )}{9}

x = arcsin (\frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (\frac{8 m - 2}{3}) + 2 πn

Solution steps

sin^{2} (x) = \frac{( 8 m - 2 )}{9}

Solve by substitution

sin^{2} (x) = \frac{8 m - 2}{9}

Let:

sin (x) = u

u^{2} = \frac{8 m - 2}{9}

u^{2} = \frac{8 m - 2}{9} : u = \frac{8 m - 2}{3}, u = - \frac{8 m - 2}{3}

u^{2} = \frac{8 m - 2}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{8 m - 2}{9}, u = - \frac{8 m - 2}{9}

Simplify

\frac{8 m - 2}{9} : \frac{8 m - 2}{3}

\frac{8 m - 2}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8 m - 2}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8 m - 2}{3}

Simplify

- \frac{8 m - 2}{9} : - \frac{8 m - 2}{3}

- \frac{8 m - 2}{9}

Simplify

\frac{8 m - 2}{9} : \frac{8 m - 2}{3}

\frac{8 m - 2}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{8 m - 2}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{8 m - 2}{3}

= - \frac{8 m - 2}{3}

u = \frac{8 m - 2}{3}, u = - \frac{8 m - 2}{3}

Substitute back

u = sin (x)

sin (x) = \frac{8 m - 2}{3}, sin (x) = - \frac{8 m - 2}{3}

sin (x) = \frac{8 m - 2}{3}, sin (x) = - \frac{8 m - 2}{3}

sin (x) = \frac{8 m - 2}{3} : x = arcsin (\frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (- \frac{8 m - 2}{3}) + 2 πn

sin (x) = \frac{8 m - 2}{3}

Apply trig inverse properties

sin (x) = \frac{8 m - 2}{3}

General solutions for

sin (x) = \frac{8 m - 2}{3}

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

x = arcsin (\frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (- \frac{8 m - 2}{3}) + 2 πn

x = arcsin (\frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (- \frac{8 m - 2}{3}) + 2 πn

sin (x) = - \frac{8 m - 2}{3} : x = arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (\frac{8 m - 2}{3}) + 2 πn

sin (x) = - \frac{8 m - 2}{3}

Apply trig inverse properties

sin (x) = - \frac{8 m - 2}{3}

General solutions for

sin (x) = - \frac{8 m - 2}{3}

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

x = arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (\frac{8 m - 2}{3}) + 2 πn

x = arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (\frac{8 m - 2}{3}) + 2 πn

Combine all the solutions

x = arcsin (\frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = arcsin (- \frac{8 m - 2}{3}) + 2 πn, x = π + arcsin (\frac{8 m - 2}{3}) + 2 πn

0.06 = 0.1 (cos (4 (t) - 1.57))

sin (x) = \frac{13}{14}

sin (x - \frac{π}{2}) + sin (\frac{π}{2}) = sin (x + \frac{π}{2})

sec^{2} (x) = 2 + sec (x)

cos (y) = 0.8192

What is the general solution for sin^2(x)=((8m-2))/9 ?
The general solution for sin^2(x)=((8m-2))/9 is