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受欢迎的三角函数 >

sin(pi/x)=0

解答

sin (\frac{π}{x}) = 0

解答

x = \frac{1}{2 n}, x = \frac{1}{1 + 2 n}; n \neq = 0, n \neq = - \frac{1}{2}

+1

度数

x = 0^{\circ} + 28.64788 \dots^{\circ} n, x = 0^{\circ} + 19.09859 \dots^{\circ} n

求解步骤

sin (\frac{π}{x}) = 0

sin (\frac{π}{x}) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{π}{x} = 0 + 2 πn, \frac{π}{x} = π + 2 πn

\frac{π}{x} = 0 + 2 πn, \frac{π}{x} = π + 2 πn

解

\frac{π}{x} = 0 + 2 πn : x = \frac{1}{2 n}; n \neq = 0

\frac{π}{x} = 0 + 2 πn

在两边乘以

x

\frac{π}{x} = 0 + 2 πn

在两边乘以

x

\frac{π}{x} x = 0 \cdot x + 2 πn x

化简

π = 0 + 2 πn x

化简

0 + 2 πn x : 2 πn x

0 + 2 πn x

0 + 2 πn x = 2 πn x

= 2 πn x

π = 2 πn x

π = 2 πn x

交换两边

2 πn x = π

两边除以

2 πn; n \neq = 0

2 πn x = π

两边除以

2 πn; n \neq = 0

\frac{2 πn x}{2 πn} = \frac{π}{2 πn}; n \neq = 0

化简

x = \frac{1}{2 n}; n \neq = 0

x = \frac{1}{2 n}; n \neq = 0

解

\frac{π}{x} = π + 2 πn : x = \frac{1}{1 + 2 n}; n \neq = - \frac{1}{2}

\frac{π}{x} = π + 2 πn

在两边乘以

x

\frac{π}{x} = π + 2 πn

在两边乘以

x

\frac{π}{x} x = π x + 2 πn x

化简

π = π x + 2 πn x

π = π x + 2 πn x

交换两边

π x + 2 πn x = π

分解

π x + 2 πn x : π x (1 + 2 n)

π x + 2 πn x

因式分解出通项

x π

= x π (1 + 2 n)

π x (1 + 2 n) = π

两边除以

π (1 + 2 n); n \neq = - \frac{1}{2}

π x (1 + 2 n) = π

两边除以

π (1 + 2 n); n \neq = - \frac{1}{2}

\frac{π x ( 1 + 2 n )}{π ( 1 + 2 n )} = \frac{π}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

化简

x = \frac{1}{1 + 2 n}; n \neq = - \frac{1}{2}

x = \frac{1}{1 + 2 n}; n \neq = - \frac{1}{2}

x = \frac{1}{2 n}, x = \frac{1}{1 + 2 n}; n \neq = 0, n \neq = - \frac{1}{2}

作图

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