What is the general solution for solvefor x,k=((1+sin(x)))/n ?

The general solution for solvefor x,k=((1+sin(x)))/n is x=arcsin(kn-1)+2pim,x=pi+arcsin(-kn+1)+2pim

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Solve for x k=(1+sin(x))/n

solve for

x, k = \frac{( 1 + sin ( x ) )}{n}

x = arcsin (kn - 1) + 2 πm, x = π + arcsin (- kn + 1) + 2 πm

Solution steps

k = \frac{( 1 + sin ( x ) )}{n}

Switch sides

\frac{1 + sin ( x )}{n} = k

Multiply both sides by

n

\frac{1 + sin ( x )}{n} = k

Multiply both sides by

n

\frac{( 1 + sin ( x ) ) n}{n} = kn; n \neq = 0

Simplify

1 + sin (x) = kn; n \neq = 0

1 + sin (x) = kn; n \neq = 0

Move

1

to the right side

1 + sin (x) = kn; n \neq = 0

Subtract

1

from both sides

1 + sin (x) - 1 = kn - 1; n \neq = 0

Simplify

sin (x) = kn - 1; n \neq = 0

sin (x) = kn - 1; n \neq = 0

Apply trig inverse properties

sin (x) = kn - 1

General solutions for

sin (x) = kn - 1

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

x = arcsin (kn - 1) + 2 πm, x = π + arcsin (- kn + 1) + 2 πm

x = arcsin (kn - 1) + 2 πm, x = π + arcsin (- kn + 1) + 2 πm

a, tan (a) = \frac{1}{3}

cos (\frac{( 4 π )}{3})

cos (\frac{( 3 π )}{4})

sec (\frac{( 5 π )}{6})

a, tan (a) = \frac{1}{3}

What is the general solution for solvefor x,k=((1+sin(x)))/n ?
The general solution for solvefor x,k=((1+sin(x)))/n is x=arcsin(kn-1)+2pim,x=pi+arcsin(-kn+1)+2pim