What is the general solution for solvefor y,sin(x)=x(1+tan(y)) ?

The general solution for solvefor y,sin(x)=x(1+tan(y)) is y=arctan((sin(x))/x-1)+pin

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solvefor y,sin(x)=x(1+tan(y))

solve for

y, sin (x) = x (1 + tan (y))

y = arctan (\frac{sin ( x )}{x} - 1) + πn

Solution steps

sin (x) = x (1 + tan (y))

Switch sides

x (1 + tan (y)) = sin (x)

Divide both sides by

x; x \neq = 0

x (1 + tan (y)) = sin (x)

Divide both sides by

x; x \neq = 0

\frac{x ( 1 + tan ( y ) )}{x} = \frac{sin ( x )}{x}; x \neq = 0

Simplify

1 + tan (y) = \frac{sin ( x )}{x}; x \neq = 0

1 + tan (y) = \frac{sin ( x )}{x}; x \neq = 0

Move

1

to the right side

1 + tan (y) = \frac{sin ( x )}{x}; x \neq = 0

Subtract

1

from both sides

1 + tan (y) - 1 = \frac{sin ( x )}{x} - 1; x \neq = 0

Simplify

tan (y) = \frac{sin ( x )}{x} - 1; x \neq = 0

tan (y) = \frac{sin ( x )}{x} - 1; x \neq = 0

Apply trig inverse properties

tan (y) = \frac{sin ( x )}{x} - 1

General solutions for

tan (y) = \frac{sin ( x )}{x} - 1

tan (x) = a \Rightarrow x = arctan (a) + πn

y = arctan (\frac{sin ( x )}{x} - 1) + πn

y = arctan (\frac{sin ( x )}{x} - 1) + πn

sin (2 x) - sin (x) = 0, (0, 2 π)

sin^{4} (x) + cos^{4} (x) = \frac{7}{9}

tan (\frac{θ}{5}) + 3 = 0

3 tan (x) - 3 = 0, 0 \leq x < 2 π

sec^{2} (θ) = 2

What is the general solution for solvefor y,sin(x)=x(1+tan(y)) ?
The general solution for solvefor y,sin(x)=x(1+tan(y)) is y=arctan((sin(x))/x-1)+pin