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受欢迎的三角函数 >

solvefor x,y+sin(xy)=1

解答

求解

x, y + sin (x y) = 1

解答

x = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}, x = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}

求解步骤

y + sin (x y) = 1

将

y

到右边

y + sin (x y) = 1

两边减去

y

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

化简

sin (x y) = 1 - y

sin (x y) = 1 - y

使用反三角函数性质

sin (x y) = 1 - y

sin (x y) = 1 - y

的通解

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

x y = arcsin (1 - y) + 2 πn, x y = π + arcsin (- 1 + y) + 2 πn

x y = arcsin (1 - y) + 2 πn, x y = π + arcsin (- 1 + y) + 2 πn

解

x y = arcsin (1 - y) + 2 πn : x = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}; y \neq = 0

x y = arcsin (1 - y) + 2 πn

两边除以

y; y \neq = 0

x y = arcsin (1 - y) + 2 πn

两边除以

y; y \neq = 0

\frac{x y}{y} = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}; y \neq = 0

化简

x = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}; y \neq = 0

x = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}; y \neq = 0

解

x y = π + arcsin (- 1 + y) + 2 πn : x = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}; y \neq = 0

x y = π + arcsin (- 1 + y) + 2 πn

两边除以

y; y \neq = 0

x y = π + arcsin (- 1 + y) + 2 πn

两边除以

y; y \neq = 0

\frac{x y}{y} = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}; y \neq = 0

化简

x = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}; y \neq = 0

x = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}; y \neq = 0

x = \frac{arcsin ( 1 - y )}{y} + \frac{2 πn}{y}, x = \frac{π}{y} + \frac{arcsin ( - 1 + y )}{y} + \frac{2 πn}{y}

作图

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