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受欢迎的三角函数 >

1/2 = 1/2*cos(4.2t+pi/6)

解答

\frac{1}{2} = \frac{1}{2} \cdot cos (4.2 t + \frac{π}{6})

解答

t = \frac{10 πn}{21} - \frac{5 π}{126}

+1

度数

t = - 7.14285 \dots^{\circ} + 85.71428 \dots^{\circ} n

求解步骤

\frac{1}{2} = \frac{1}{2} cos (4.2 t + \frac{π}{6})

交换两边

\frac{1}{2} cos (4.2 t + \frac{π}{6}) = \frac{1}{2}

在两边乘以

2

\frac{1}{2} cos (4.2 t + \frac{π}{6}) = \frac{1}{2}

在两边乘以

2

2 \cdot \frac{1}{2} cos (4.2 t + \frac{π}{6}) = \frac{1 \cdot 2}{2}

化简

2 \cdot \frac{1}{2} cos (4.2 t + \frac{π}{6}) = \frac{1 \cdot 2}{2}

化简

2 \cdot \frac{1}{2} cos (4.2 t + \frac{π}{6}) : cos (4.2 t + \frac{π}{6})

2 \cdot \frac{1}{2} cos (4.2 t + \frac{π}{6})

分式相乘:

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

= \frac{1 \cdot 2}{2} cos (4.2 t + \frac{π}{6})

约分:

2

= cos (4.2 t + \frac{π}{6}) \cdot 1

乘以:

cos (4.2 t + \frac{π}{6}) \cdot 1 = cos (4.2 t + \frac{π}{6})

= cos (4.2 t + \frac{π}{6})

化简

\frac{1 \cdot 2}{2} : 1

\frac{1 \cdot 2}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

cos (4.2 t + \frac{π}{6}) = 1

cos (4.2 t + \frac{π}{6}) = 1

cos (4.2 t + \frac{π}{6}) = 1

cos (4.2 t + \frac{π}{6}) = 1

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}

4.2 t + \frac{π}{6} = 0 + 2 πn

4.2 t + \frac{π}{6} = 0 + 2 πn

解

4.2 t + \frac{π}{6} = 0 + 2 πn : t = \frac{10 πn}{21} - \frac{5 π}{126}

4.2 t + \frac{π}{6} = 0 + 2 πn

0 + 2 πn = 2 πn

4.2 t + \frac{π}{6} = 2 πn

在两边乘以

10

4.2 t + \frac{π}{6} = 2 πn

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

4.2 t \cdot 10 + \frac{π}{6} \cdot 10 = 2 πn \cdot 10

整理后得

42 t + \frac{5 π}{3} = 20 πn

42 t + \frac{5 π}{3} = 20 πn

将

\frac{5 π}{3}

到右边

42 t + \frac{5 π}{3} = 20 πn

两边减去

\frac{5 π}{3}

42 t + \frac{5 π}{3} - \frac{5 π}{3} = 20 πn - \frac{5 π}{3}

化简

42 t = 20 πn - \frac{5 π}{3}

42 t = 20 πn - \frac{5 π}{3}

两边除以

42

42 t = 20 πn - \frac{5 π}{3}

两边除以

42

\frac{42 t}{42} = \frac{20 πn}{42} - \frac{\frac{5 π}{3}}{42}

化简

\frac{42 t}{42} = \frac{20 πn}{42} - \frac{\frac{5 π}{3}}{42}

化简

\frac{42 t}{42} : t

\frac{42 t}{42}

数字相除:

\frac{42}{42} = 1

= t

化简

\frac{20 πn}{42} - \frac{\frac{5 π}{3}}{42} : \frac{10 πn}{21} - \frac{5 π}{126}

\frac{20 πn}{42} - \frac{\frac{5 π}{3}}{42}

\frac{20 πn}{42} = \frac{10 πn}{21}

\frac{20 πn}{42}

约分:

2

= \frac{10 πn}{21}

\frac{\frac{5 π}{3}}{42} = \frac{5 π}{126}

\frac{\frac{5 π}{3}}{42}

使用分式法则:

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

= \frac{5 π}{3 \cdot 42}

数字相乘:

3 \cdot 42 = 126

= \frac{5 π}{126}

= \frac{10 πn}{21} - \frac{5 π}{126}

t = \frac{10 πn}{21} - \frac{5 π}{126}

t = \frac{10 πn}{21} - \frac{5 π}{126}

t = \frac{10 πn}{21} - \frac{5 π}{126}

t = \frac{10 πn}{21} - \frac{5 π}{126}

作图

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