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受欢迎的三角函数 >

tan(2x)=3sin(2x)

解答

tan (2 x) = 3 sin (2 x)

解答

x = πn, x = \frac{π}{2} + πn, x = \frac{1.23095 \dots}{2} + πn, x = π - \frac{1.23095 \dots}{2} + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n, x = 35.26438 \dots^{\circ} + 18 0^{\circ} n, x = 144.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 x) = 3 sin (2 x)

两边减去

3 sin (2 x)

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

用 sin, cos 表示

tan (2 x) - 3 sin (2 x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 2 x )}{cos ( 2 x )} - 3 sin (2 x)

化简

\frac{sin ( 2 x )}{cos ( 2 x )} - 3 sin (2 x) : \frac{sin ( 2 x ) - 3 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x )}{cos ( 2 x )} - 3 sin (2 x)

将项转换为分式:

3 sin (2 x) = \frac{3 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x )}{cos ( 2 x )} - \frac{3 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

因为分母相等，所以合并分式:

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

= \frac{sin ( 2 x ) - 3 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) - 3 sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 x) - 3 cos (2 x) sin (2 x) = 0

分解

sin (2 x) - 3 cos (2 x) sin (2 x) : - sin (2 x) (3 cos (2 x) - 1)

sin (2 x) - 3 cos (2 x) sin (2 x)

因式分解出通项

- sin (2 x)

= - sin (2 x) (- 1 + 3 cos (2 x))

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

分别求解每个部分

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

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

sin (2 x) = 0

sin (2 x) = 0

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 = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

解

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

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

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

3 cos (2 x) - 1 = 0 : x = \frac{arccos ( \frac{1}{3} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

3 cos (2 x) - 1 = 0

将

1

到右边

3 cos (2 x) - 1 = 0

两边加上

1

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

化简

3 cos (2 x) = 1

3 cos (2 x) = 1

两边除以

3

3 cos (2 x) = 1

两边除以

3

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

化简

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

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

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (\frac{1}{3}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

2 x = arccos (\frac{1}{3}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

解

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

2 x = arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 x = arccos (\frac{1}{3}) + 2 πn

两边除以

2

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

化简

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

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

解

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn : x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

x = \frac{arccos ( \frac{1}{3} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

合并所有解

x = πn, x = \frac{π}{2} + πn, x = \frac{arccos ( \frac{1}{3} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

以小数形式表示解

x = πn, x = \frac{π}{2} + πn, x = \frac{1.23095 \dots}{2} + πn, x = π - \frac{1.23095 \dots}{2} + πn

作图

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