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受欢迎的三角函数 >

(2sin(2x)-1)(3tan(2x)-1)=0

解答

(2 sin (2 x) - 1) (3 tan (2 x) - 1) = 0

解答

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = \frac{0.32175 \dots}{2} + \frac{πn}{2}

+1

度数

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n, x = 9.21747 \dots^{\circ} + 9 0^{\circ} n

求解步骤

(2 sin (2 x) - 1) (3 tan (2 x) - 1) = 0

分别求解每个部分

2 sin (2 x) - 1 = 0 or 3 tan (2 x) - 1 = 0

2 sin (2 x) - 1 = 0 : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

2 sin (2 x) - 1 = 0

将

1

到右边

2 sin (2 x) - 1 = 0

两边加上

1

2 sin (2 x) - 1 + 1 = 0 + 1

化简

2 sin (2 x) = 1

2 sin (2 x) = 1

两边除以

2

2 sin (2 x) = 1

两边除以

2

\frac{2 sin ( 2 x )}{2} = \frac{1}{2}

化简

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

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}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

解

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

3 tan (2 x) - 1 = 0 : x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

3 tan (2 x) - 1 = 0

将

1

到右边

3 tan (2 x) - 1 = 0

两边加上

1

3 tan (2 x) - 1 + 1 = 0 + 1

化简

3 tan (2 x) = 1

3 tan (2 x) = 1

两边除以

3

3 tan (2 x) = 1

两边除以

3

\frac{3 tan ( 2 x )}{3} = \frac{1}{3}

化简

tan (2 x) = \frac{1}{3}

tan (2 x) = \frac{1}{3}

使用反三角函数性质

tan (2 x) = \frac{1}{3}

tan (2 x) = \frac{1}{3}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

2 x = arctan (\frac{1}{3}) + πn

2 x = arctan (\frac{1}{3}) + πn

解

2 x = arctan (\frac{1}{3}) + πn : x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

2 x = arctan (\frac{1}{3}) + πn

两边除以

2

2 x = arctan (\frac{1}{3}) + πn

两边除以

2

\frac{2 x}{2} = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

化简

x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

合并所有解

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = \frac{arctan ( \frac{1}{3} )}{2} + \frac{πn}{2}

以小数形式表示解

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn, x = \frac{0.32175 \dots}{2} + \frac{πn}{2}

作图

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