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受欢迎的三角函数 >

3sin(2x+30)=tan(2x+30)

解答

3 sin (2 x + 30) = tan (2 x + 30)

解答

x = πn - 15, x = \frac{π}{2} + πn - 15, x = \frac{1.23095 \dots}{2} + πn - 15, x = π - \frac{1.23095 \dots}{2} + πn - 15

+1

度数

x = - 859.43669 \dots^{\circ} + 18 0^{\circ} n, x = - 769.43669 \dots^{\circ} + 18 0^{\circ} n, x = - 824.17230 \dots^{\circ} + 18 0^{\circ} n, x = - 714.70108 \dots^{\circ} + 18 0^{\circ} n

求解步骤

3 sin (2 x + 30) = tan (2 x + 30)

两边减去

tan (2 x + 30)

3 sin (2 x + 30) - tan (2 x + 30) = 0

用 sin, cos 表示

- tan (30 + 2 x) + 3 sin (30 + 2 x)

使用基本三角恒等式:

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

= - \frac{sin ( 30 + 2 x )}{cos ( 30 + 2 x )} + 3 sin (30 + 2 x)

化简

- \frac{sin ( 30 + 2 x )}{cos ( 30 + 2 x )} + 3 sin (30 + 2 x) : \frac{- sin ( 30 + 2 x ) + 3 sin ( 30 + 2 x ) cos ( 30 + 2 x )}{cos ( 30 + 2 x )}

- \frac{sin ( 30 + 2 x )}{cos ( 30 + 2 x )} + 3 sin (30 + 2 x)

将项转换为分式:

3 sin (2 x + 30) = \frac{3 sin ( 30 + 2 x ) cos ( 30 + 2 x )}{cos ( 30 + 2 x )}

= - \frac{sin ( 30 + 2 x )}{cos ( 30 + 2 x )} + \frac{3 sin ( 30 + 2 x ) cos ( 30 + 2 x )}{cos ( 30 + 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 ( 30 + 2 x ) + 3 sin ( 30 + 2 x ) cos ( 30 + 2 x )}{cos ( 30 + 2 x )}

= \frac{- sin ( 30 + 2 x ) + 3 sin ( 30 + 2 x ) cos ( 30 + 2 x )}{cos ( 30 + 2 x )}

\frac{- sin ( 30 + 2 x ) + 3 cos ( 30 + 2 x ) sin ( 30 + 2 x )}{cos ( 30 + 2 x )} = 0

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

- sin (30 + 2 x) + 3 cos (30 + 2 x) sin (30 + 2 x) = 0

分解

- sin (30 + 2 x) + 3 cos (30 + 2 x) sin (30 + 2 x) : sin (2 (x + 15)) (3 cos (2 (x + 15)) - 1)

- sin (30 + 2 x) + 3 cos (30 + 2 x) sin (30 + 2 x)

因式分解出通项

sin (30 + 2 x)

= sin (30 + 2 x) (- 1 + 3 cos (30 + 2 x))

分解

2 x + 30 : 2 (x + 15)

2 x + 30

因式分解出通项

2 : 2 (x + 15)

2 x + 30

将

30

改写为

2 \cdot 15

= 2 x + 2 \cdot 15

因式分解出通项

2

= 2 (x + 15)

= 2 (x + 15)

= sin (2 x + 30) (3 cos (2 (x + 15)) - 1)

分解

2 x + 30 : 2 (x + 15)

2 x + 30

因式分解出通项

2 : 2 (x + 15)

2 x + 30

将

30

改写为

2 \cdot 15

= 2 x + 2 \cdot 15

因式分解出通项

2

= 2 (x + 15)

= 2 (x + 15)

= sin (2 (x + 15)) (3 cos (2 (x + 15)) - 1)

sin (2 (x + 15)) (3 cos (2 (x + 15)) - 1) = 0

分别求解每个部分

sin (2 (x + 15)) = 0 or 3 cos (2 (x + 15)) - 1 = 0

sin (2 (x + 15)) = 0 : x = πn - 15, x = \frac{π}{2} + πn - 15

sin (2 (x + 15)) = 0

sin (2 (x + 15)) = 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 + 15) = 0 + 2 πn, 2 (x + 15) = π + 2 πn

2 (x + 15) = 0 + 2 πn, 2 (x + 15) = π + 2 πn

解

2 (x + 15) = 0 + 2 πn : x = πn - 15

2 (x + 15) = 0 + 2 πn

0 + 2 πn = 2 πn

2 (x + 15) = 2 πn

两边除以

2

2 (x + 15) = 2 πn

两边除以

2

\frac{2 ( x + 15 )}{2} = \frac{2 πn}{2}

化简

x + 15 = πn

x + 15 = πn

将

15

到右边

x + 15 = πn

两边减去

15

x + 15 - 15 = πn - 15

化简

x = πn - 15

x = πn - 15

解

2 (x + 15) = π + 2 πn : x = \frac{π}{2} + πn - 15

2 (x + 15) = π + 2 πn

两边除以

2

2 (x + 15) = π + 2 πn

两边除以

2

\frac{2 ( x + 15 )}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

x + 15 = \frac{π}{2} + πn

x + 15 = \frac{π}{2} + πn

将

15

到右边

x + 15 = \frac{π}{2} + πn

两边减去

15

x + 15 - 15 = \frac{π}{2} + πn - 15

化简

x = \frac{π}{2} + πn - 15

x = \frac{π}{2} + πn - 15

x = πn - 15, x = \frac{π}{2} + πn - 15

3 cos (2 (x + 15)) - 1 = 0 : x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

3 cos (2 (x + 15)) - 1 = 0

将

1

到右边

3 cos (2 (x + 15)) - 1 = 0

两边加上

1

3 cos (2 (x + 15)) - 1 + 1 = 0 + 1

化简

3 cos (2 (x + 15)) = 1

3 cos (2 (x + 15)) = 1

两边除以

3

3 cos (2 (x + 15)) = 1

两边除以

3

\frac{3 cos ( 2 ( x + 15 ) )}{3} = \frac{1}{3}

化简

cos (2 (x + 15)) = \frac{1}{3}

cos (2 (x + 15)) = \frac{1}{3}

使用反三角函数性质

cos (2 (x + 15)) = \frac{1}{3}

cos (2 (x + 15)) = \frac{1}{3}

的通解

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

2 (x + 15) = arccos (\frac{1}{3}) + 2 πn, 2 (x + 15) = 2 π - arccos (\frac{1}{3}) + 2 πn

2 (x + 15) = arccos (\frac{1}{3}) + 2 πn, 2 (x + 15) = 2 π - arccos (\frac{1}{3}) + 2 πn

解

2 (x + 15) = arccos (\frac{1}{3}) + 2 πn : x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

2 (x + 15) = arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 (x + 15) = arccos (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 ( x + 15 )}{2} = \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

x + 15 = \frac{arccos ( \frac{1}{3} )}{2} + πn

x + 15 = \frac{arccos ( \frac{1}{3} )}{2} + πn

将

15

到右边

x + 15 = \frac{arccos ( \frac{1}{3} )}{2} + πn

两边减去

15

x + 15 - 15 = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

化简

x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

解

2 (x + 15) = 2 π - arccos (\frac{1}{3}) + 2 πn : x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

2 (x + 15) = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

2 (x + 15) = 2 π - arccos (\frac{1}{3}) + 2 πn

两边除以

2

\frac{2 ( x + 15 )}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

x + 15 = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

x + 15 = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

将

15

到右边

x + 15 = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

两边减去

15

x + 15 - 15 = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

化简

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

合并所有解

x = πn - 15, x = \frac{π}{2} + πn - 15, x = \frac{arccos ( \frac{1}{3} )}{2} + πn - 15, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn - 15

以小数形式表示解

x = πn - 15, x = \frac{π}{2} + πn - 15, x = \frac{1.23095 \dots}{2} + πn - 15, x = π - \frac{1.23095 \dots}{2} + πn - 15

作图

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