What is the general solution for solvefor x,y=arctan(2x+3y) ?

The general solution for solvefor x,y=arctan(2x+3y) is x=(tan(y))/2-(3y)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,y=arctan(2x+3y)

solve for

x, y = arctan (2 x + 3 y)

x = \frac{tan ( y )}{2} - \frac{3 y}{2}

Solution steps

y = arctan (2 x + 3 y)

Switch sides

arctan (2 x + 3 y) = y

Apply trig inverse properties

arctan (2 x + 3 y) = y

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

2 x + 3 y = tan (y)

2 x + 3 y = tan (y)

Solve

2 x + 3 y = tan (y) : x = \frac{tan ( y )}{2} - \frac{3 y}{2}

2 x + 3 y = tan (y)

Move

3 y

to the right side

2 x + 3 y = tan (y)

Subtract

3 y

from both sides

2 x + 3 y - 3 y = tan (y) - 3 y

Simplify

2 x = tan (y) - 3 y

2 x = tan (y) - 3 y

Divide both sides by

2

2 x = tan (y) - 3 y

Divide both sides by

2

\frac{2 x}{2} = \frac{tan ( y )}{2} - \frac{3 y}{2}

Simplify

x = \frac{tan ( y )}{2} - \frac{3 y}{2}

x = \frac{tan ( y )}{2} - \frac{3 y}{2}

x = \frac{tan ( y )}{2} - \frac{3 y}{2}

sin (2 x) = 0.89583

tan (θ) = 1.234, θ

x, sin (x) cos (x) = sin (x + p) cos (x + p)

sin (θ) = \frac{3}{5}, cos (2 θ)

1 1^{2} = 6^{2} + 7^{2} - 2 \cdot 6 \cdot 7 \cdot cos (x)

What is the general solution for solvefor x,y=arctan(2x+3y) ?
The general solution for solvefor x,y=arctan(2x+3y) is x=(tan(y))/2-(3y)/2