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受欢迎的三角函数 >

sin((3pi)/x)=0

解答

sin (\frac{3 π}{x}) = 0

解答

x = \frac{3}{2 n}, x = \frac{3}{1 + 2 n}; n \neq = 0, n \neq = - \frac{1}{2}

+1

度数

x = 0^{\circ} + 85.94366 \dots^{\circ} n, x = 0^{\circ} + 57.29577 \dots^{\circ} n

求解步骤

sin (\frac{3 π}{x}) = 0

sin (\frac{3 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{3 π}{x} = 0 + 2 πn, \frac{3 π}{x} = π + 2 πn

\frac{3 π}{x} = 0 + 2 πn, \frac{3 π}{x} = π + 2 πn

解

\frac{3 π}{x} = 0 + 2 πn : x = \frac{3}{2 n}; n \neq = 0

\frac{3 π}{x} = 0 + 2 πn

在两边乘以

x

\frac{3 π}{x} = 0 + 2 πn

在两边乘以

x

\frac{3 π}{x} x = 0 \cdot x + 2 πn x

化简

3 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

3 π = 2 πn x

3 π = 2 πn x

交换两边

2 πn x = 3 π

两边除以

2 πn; n \neq = 0

2 πn x = 3 π

两边除以

2 πn; n \neq = 0

\frac{2 πn x}{2 πn} = \frac{3 π}{2 πn}; n \neq = 0

化简

x = \frac{3}{2 n}; n \neq = 0

x = \frac{3}{2 n}; n \neq = 0

解

\frac{3 π}{x} = π + 2 πn : x = \frac{3}{1 + 2 n}; n \neq = - \frac{1}{2}

\frac{3 π}{x} = π + 2 πn

在两边乘以

x

\frac{3 π}{x} = π + 2 πn

在两边乘以

x

\frac{3 π}{x} x = π x + 2 πn x

化简

3 π = π x + 2 πn x

3 π = π x + 2 πn x

交换两边

π x + 2 πn x = 3 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) = 3 π

两边除以

π (1 + 2 n); n \neq = - \frac{1}{2}

π x (1 + 2 n) = 3 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{3 π}{π ( 1 + 2 n )}; n \neq = - \frac{1}{2}

化简

x = \frac{3}{1 + 2 n}; n \neq = - \frac{1}{2}

x = \frac{3}{1 + 2 n}; n \neq = - \frac{1}{2}

x = \frac{3}{2 n}, x = \frac{3}{1 + 2 n}; n \neq = 0, n \neq = - \frac{1}{2}

作图

流行的例子

solvefor x,(sin(sqrt(x))-log_{3}(12))=20

so l v e f or x, (sin (x) - lo g_{3} (12)) = 20

tan(θ)=-(sqrt(7))/5 , pi/2 <θ<pi,cos(2θ)

tan (θ) = - \frac{7}{5}, \frac{π}{2} < θ < π, cos (2 θ)

cos(θ)=-(1/(sqrt(2)))

cos (θ) = - (\frac{1}{2})

sqrt(cos(θ))=2cos(θ)-1

cos (θ) = 2 cos (θ) - 1

2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}