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受欢迎的三角函数 >

cos(0.5x)=0.5,-pi<= x<= pi

解答

cos (0.5 x) = 0.5, - π \leq x \leq π

解答

x = \frac{2 π}{3}, x = - \frac{2 π}{3}

度数

x = 12 0^{\circ}, x = - 12 0^{\circ}

求解步骤

cos (0.5 x) = 0.5, - π \leq x \leq π

cos (0.5 x) = 0.5

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}

0.5 x = \frac{π}{3} + 2 πn, 0.5 x = \frac{5 π}{3} + 2 πn

0.5 x = \frac{π}{3} + 2 πn, 0.5 x = \frac{5 π}{3} + 2 πn

解

0.5 x = \frac{π}{3} + 2 πn : x = \frac{2 π}{3} + 4 πn

0.5 x = \frac{π}{3} + 2 πn

在两边乘以

10

0.5 x = \frac{π}{3} + 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

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

整理后得

5 x = \frac{10 π}{3} + 20 πn

5 x = \frac{10 π}{3} + 20 πn

两边除以

5

5 x = \frac{10 π}{3} + 20 πn

两边除以

5

\frac{5 x}{5} = \frac{\frac{10 π}{3}}{5} + \frac{20 πn}{5}

化简

\frac{5 x}{5} = \frac{\frac{10 π}{3}}{5} + \frac{20 πn}{5}

化简

\frac{5 x}{5} : x

\frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

\frac{\frac{10 π}{3}}{5} + \frac{20 πn}{5} : \frac{2 π}{3} + 4 πn

\frac{\frac{10 π}{3}}{5} + \frac{20 πn}{5}

\frac{\frac{10 π}{3}}{5} = \frac{2 π}{3}

\frac{\frac{10 π}{3}}{5}

使用分式法则:

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

= \frac{10 π}{3 \cdot 5}

数字相乘:

3 \cdot 5 = 15

= \frac{10 π}{15}

约分:

5

= \frac{2 π}{3}

\frac{20 πn}{5} = 4 πn

\frac{20 πn}{5}

数字相除:

\frac{20}{5} = 4

= 4 πn

= \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

解

0.5 x = \frac{5 π}{3} + 2 πn : x = \frac{10 π}{3} + 4 πn

0.5 x = \frac{5 π}{3} + 2 πn

在两边乘以

10

0.5 x = \frac{5 π}{3} + 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

0.5 x \cdot 10 = \frac{5 π}{3} \cdot 10 + 2 πn \cdot 10

整理后得

5 x = \frac{50 π}{3} + 20 πn

5 x = \frac{50 π}{3} + 20 πn

两边除以

5

5 x = \frac{50 π}{3} + 20 πn

两边除以

5

\frac{5 x}{5} = \frac{\frac{50 π}{3}}{5} + \frac{20 πn}{5}

化简

\frac{5 x}{5} = \frac{\frac{50 π}{3}}{5} + \frac{20 πn}{5}

化简

\frac{5 x}{5} : x

\frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

\frac{\frac{50 π}{3}}{5} + \frac{20 πn}{5} : \frac{10 π}{3} + 4 πn

\frac{\frac{50 π}{3}}{5} + \frac{20 πn}{5}

\frac{\frac{50 π}{3}}{5} = \frac{10 π}{3}

\frac{\frac{50 π}{3}}{5}

使用分式法则:

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

= \frac{50 π}{3 \cdot 5}

数字相乘:

3 \cdot 5 = 15

= \frac{50 π}{15}

约分:

5

= \frac{10 π}{3}

\frac{20 πn}{5} = 4 πn

\frac{20 πn}{5}

数字相除:

\frac{20}{5} = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

在

- π \leq x \leq π

范围内的解

x = \frac{2 π}{3}, x = - \frac{2 π}{3}

作图

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