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受欢迎的三角函数 >

0=(2pi^2cos(\frac{pix)/(10))}{25}

解答

0 = \frac{2 π ^{2} cos ( \frac{π x}{10} )}{25}

解答

x = 5 + 20 n, x = 15 + 20 n

+1

度数

x = 286.47889 \dots^{\circ} + 1145.91559 \dots^{\circ} n, x = 859.43669 \dots^{\circ} + 1145.91559 \dots^{\circ} n

求解步骤

0 = \frac{2 π ^{2} cos ( \frac{π x}{10} )}{25}

交换两边

\frac{2 π ^{2} cos ( \frac{π x}{10} )}{25} = 0

在两边乘以

25

\frac{2 π ^{2} cos ( \frac{π x}{10} )}{25} = 0

在两边乘以

25

\frac{25 \cdot 2 π ^{2} cos ( \frac{π x}{10} )}{25} = 0 \cdot 25

化简

2 π^{2} cos (\frac{π x}{10}) = 0

2 π^{2} cos (\frac{π x}{10}) = 0

两边除以

2 π^{2}

2 π^{2} cos (\frac{π x}{10}) = 0

两边除以

2 π^{2}

\frac{2 π ^{2} cos ( \frac{π x}{10} )}{2 π ^{2}} = \frac{0}{2 π ^{2}}

化简

cos (\frac{π x}{10}) = 0

cos (\frac{π x}{10}) = 0

cos (\frac{π x}{10}) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{π x}{10} = \frac{π}{2} + 2 πn, \frac{π x}{10} = \frac{3 π}{2} + 2 πn

\frac{π x}{10} = \frac{π}{2} + 2 πn, \frac{π x}{10} = \frac{3 π}{2} + 2 πn

解

\frac{π x}{10} = \frac{π}{2} + 2 πn : x = 5 + 20 n

\frac{π x}{10} = \frac{π}{2} + 2 πn

在两边乘以

10

\frac{π x}{10} = \frac{π}{2} + 2 πn

在两边乘以

10

\frac{10 π x}{10} = 10 \cdot \frac{π}{2} + 10 \cdot 2 πn

化简

\frac{10 π x}{10} = 10 \cdot \frac{π}{2} + 10 \cdot 2 πn

化简

\frac{10 π x}{10} : π x

\frac{10 π x}{10}

数字相除:

\frac{10}{10} = 1

= π x

化简

10 \cdot \frac{π}{2} + 10 \cdot 2 πn : 5 π + 20 πn

10 \cdot \frac{π}{2} + 10 \cdot 2 πn

10 \cdot \frac{π}{2} = 5 π

10 \cdot \frac{π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 10}{2}

数字相除:

\frac{10}{2} = 5

= 5 π

10 \cdot 2 πn = 20 πn

10 \cdot 2 πn

数字相乘:

10 \cdot 2 = 20

= 20 πn

= 5 π + 20 πn

π x = 5 π + 20 πn

π x = 5 π + 20 πn

π x = 5 π + 20 πn

两边除以

π

π x = 5 π + 20 πn

两边除以

π

\frac{π x}{π} = \frac{5 π}{π} + \frac{20 πn}{π}

化简

\frac{π x}{π} = \frac{5 π}{π} + \frac{20 πn}{π}

化简

\frac{π x}{π} : x

\frac{π x}{π}

约分:

π

= x

化简

\frac{5 π}{π} + \frac{20 πn}{π} : 5 + 20 n

\frac{5 π}{π} + \frac{20 πn}{π}

消掉

\frac{5 π}{π} : 5

\frac{5 π}{π}

约分:

π

= 5

= 5 + \frac{20 πn}{π}

消掉

\frac{20 πn}{π} : 20 n

\frac{20 πn}{π}

约分:

π

= 20 n

= 5 + 20 n

x = 5 + 20 n

x = 5 + 20 n

x = 5 + 20 n

解

\frac{π x}{10} = \frac{3 π}{2} + 2 πn : x = 15 + 20 n

\frac{π x}{10} = \frac{3 π}{2} + 2 πn

在两边乘以

10

\frac{π x}{10} = \frac{3 π}{2} + 2 πn

在两边乘以

10

\frac{10 π x}{10} = 10 \cdot \frac{3 π}{2} + 10 \cdot 2 πn

化简

\frac{10 π x}{10} = 10 \cdot \frac{3 π}{2} + 10 \cdot 2 πn

化简

\frac{10 π x}{10} : π x

\frac{10 π x}{10}

数字相除:

\frac{10}{10} = 1

= π x

化简

10 \cdot \frac{3 π}{2} + 10 \cdot 2 πn : 15 π + 20 πn

10 \cdot \frac{3 π}{2} + 10 \cdot 2 πn

10 \cdot \frac{3 π}{2} = 15 π

10 \cdot \frac{3 π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 π 10}{2}

数字相乘:

3 \cdot 10 = 30

= \frac{30 π}{2}

数字相除:

\frac{30}{2} = 15

= 15 π

10 \cdot 2 πn = 20 πn

10 \cdot 2 πn

数字相乘:

10 \cdot 2 = 20

= 20 πn

= 15 π + 20 πn

π x = 15 π + 20 πn

π x = 15 π + 20 πn

π x = 15 π + 20 πn

两边除以

π

π x = 15 π + 20 πn

两边除以

π

\frac{π x}{π} = \frac{15 π}{π} + \frac{20 πn}{π}

化简

\frac{π x}{π} = \frac{15 π}{π} + \frac{20 πn}{π}

化简

\frac{π x}{π} : x

\frac{π x}{π}

约分:

π

= x

化简

\frac{15 π}{π} + \frac{20 πn}{π} : 15 + 20 n

\frac{15 π}{π} + \frac{20 πn}{π}

消掉

\frac{15 π}{π} : 15

\frac{15 π}{π}

约分:

π

= 15

= 15 + \frac{20 πn}{π}

消掉

\frac{20 πn}{π} : 20 n

\frac{20 πn}{π}

约分:

π

= 20 n

= 15 + 20 n

x = 15 + 20 n

x = 15 + 20 n

x = 15 + 20 n

x = 5 + 20 n, x = 15 + 20 n

作图

流行的例子

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300 cos (7 0^{\circ}) = 375 cos (θ)

812.25=399.6001+400-1140cos(a)

812.25 = 399.6001 + 400 - 1140 cos (a)

- 0.6 = sin (2 x)

solvefor x,(sin(pix)+cos(piy))^5=36

so l v e f or x, (sin (π x) + cos (π y))^{5} = 36

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