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受欢迎的三角函数 >

2cos(pi/2 t)=sqrt(3)

解答

2 cos (\frac{π}{2} t) = 3

解答

t = \frac{1}{3} + 4 n, t = \frac{11}{3} + 4 n

+1

度数

t = 19.09859 \dots^{\circ} + 229.18311 \dots^{\circ} n, t = 210.08452 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

2 cos (\frac{π}{2} t) = 3

两边除以

2

2 cos (\frac{π}{2} t) = 3

两边除以

2

\frac{2 cos ( \frac{π}{2} t )}{2} = \frac{3}{2}

化简

cos (\frac{π}{2} t) = \frac{3}{2}

cos (\frac{π}{2} t) = \frac{3}{2}

cos (\frac{π}{2} t) = \frac{3}{2}

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}{2} t = \frac{π}{6} + 2 πn, \frac{π}{2} t = \frac{11 π}{6} + 2 πn

\frac{π}{2} t = \frac{π}{6} + 2 πn, \frac{π}{2} t = \frac{11 π}{6} + 2 πn

解

\frac{π}{2} t = \frac{π}{6} + 2 πn : t = \frac{1}{3} + 4 n

\frac{π}{2} t = \frac{π}{6} + 2 πn

在两边乘以

2

\frac{π}{2} t = \frac{π}{6} + 2 πn

在两边乘以

2

2 \cdot \frac{π}{2} t = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} t = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} t : π t

2 \cdot \frac{π}{2} t

分式相乘:

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

= \frac{2 π}{2} t

约分:

2

= t π

化简

2 \cdot \frac{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

分式相乘:

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

= \frac{π 2}{6}

约分:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

π t = \frac{π}{3} + 4 πn

π t = \frac{π}{3} + 4 πn

π t = \frac{π}{3} + 4 πn

两边除以

π

π t = \frac{π}{3} + 4 πn

两边除以

π

\frac{π t}{π} = \frac{\frac{π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} = \frac{\frac{π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{\frac{π}{3}}{π} + \frac{4 πn}{π} : \frac{1}{3} + 4 n

\frac{\frac{π}{3}}{π} + \frac{4 πn}{π}

\frac{\frac{π}{3}}{π} = \frac{1}{3}

\frac{\frac{π}{3}}{π}

使用分式法则:

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

= \frac{π}{3 π}

约分:

π

= \frac{1}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{1}{3} + 4 n

t = \frac{1}{3} + 4 n

t = \frac{1}{3} + 4 n

t = \frac{1}{3} + 4 n

解

\frac{π}{2} t = \frac{11 π}{6} + 2 πn : t = \frac{11}{3} + 4 n

\frac{π}{2} t = \frac{11 π}{6} + 2 πn

在两边乘以

2

\frac{π}{2} t = \frac{11 π}{6} + 2 πn

在两边乘以

2

2 \cdot \frac{π}{2} t = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} t = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

化简

2 \cdot \frac{π}{2} t : π t

2 \cdot \frac{π}{2} t

分式相乘:

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

= \frac{2 π}{2} t

约分:

2

= t π

化简

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

分式相乘:

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

= \frac{11 π 2}{6}

数字相乘:

11 \cdot 2 = 22

= \frac{22 π}{6}

约分:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

π t = \frac{11 π}{3} + 4 πn

π t = \frac{11 π}{3} + 4 πn

π t = \frac{11 π}{3} + 4 πn

两边除以

π

π t = \frac{11 π}{3} + 4 πn

两边除以

π

\frac{π t}{π} = \frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} = \frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π} : \frac{11}{3} + 4 n

\frac{\frac{11 π}{3}}{π} + \frac{4 πn}{π}

\frac{\frac{11 π}{3}}{π} = \frac{11}{3}

\frac{\frac{11 π}{3}}{π}

使用分式法则:

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

= \frac{11 π}{3 π}

约分:

π

= \frac{11}{3}

\frac{4 πn}{π} = 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= \frac{11}{3} + 4 n

t = \frac{11}{3} + 4 n

t = \frac{11}{3} + 4 n

t = \frac{11}{3} + 4 n

t = \frac{1}{3} + 4 n, t = \frac{11}{3} + 4 n

作图

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