What is the general solution for solvefor y,x=tan(pi/4 y) ?

The general solution for solvefor y,x=tan(pi/4 y) is y=(4arctan(x))/pi+4n

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solvefor y,x=tan(pi/4 y)

solve for

y, x = tan (\frac{π}{4} y)

y = \frac{4 arctan ( x )}{π} + 4 n

Solution steps

x = tan (\frac{π}{4} y)

Switch sides

tan (\frac{π}{4} y) = x

Apply trig inverse properties

tan (\frac{π}{4} y) = x

General solutions for

tan (\frac{π}{4} y) = x

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

\frac{π}{4} y = arctan (x) + πn

\frac{π}{4} y = arctan (x) + πn

Solve

\frac{π}{4} y = arctan (x) + πn : y = \frac{4 arctan ( x )}{π} + 4 n

\frac{π}{4} y = arctan (x) + πn

Multiply both sides by

4

\frac{π}{4} y = arctan (x) + πn

Multiply both sides by

4

4 \cdot \frac{π}{4} y = 4 arctan (x) + 4 πn

Simplify

π y = 4 arctan (x) + 4 πn

π y = 4 arctan (x) + 4 πn

Divide both sides by

π

π y = 4 arctan (x) + 4 πn

Divide both sides by

π

\frac{π y}{π} = \frac{4 arctan ( x )}{π} + \frac{4 πn}{π}

Simplify

y = \frac{4 arctan ( x )}{π} + 4 n

y = \frac{4 arctan ( x )}{π} + 4 n

y = \frac{4 arctan ( x )}{π} + 4 n

2 sin^{2} (x) - 5 sin (x) + 2 = 0, 0 \leq x \leq 2 π

sin^{2} (x) - 4 cos (x) + 4 = 0

sec (θ) = sin^{2} (θ) sec (θ)

sin (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π

1 + cos (x) = 2 - cos (x)

What is the general solution for solvefor y,x=tan(pi/4 y) ?
The general solution for solvefor y,x=tan(pi/4 y) is y=(4arctan(x))/pi+4n