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受欢迎的三角函数 >

tan(2x+10)=cot(3x-15)

解答

tan (2 x + 10) = cot (3 x - 15)

解答

x = \frac{π}{10} + \frac{2 πn}{5} + 1, x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

+1

度数

x = 75.29577 \dots^{\circ} + 7 2^{\circ} n, x = 111.29577 \dots^{\circ} + 7 2^{\circ} n

求解步骤

tan (2 x + 10) = cot (3 x - 15)

两边减去

cot (3 x - 15)

tan (2 x + 10) - cot (3 x - 15) = 0

用 sin, cos 表示

- cot (- 15 + 3 x) + tan (10 + 2 x)

使用基本三角恒等式:

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

= - \frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} + tan (10 + 2 x)

使用基本三角恒等式:

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

= - \frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} + \frac{sin ( 10 + 2 x )}{cos ( 10 + 2 x )}

化简

- \frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} + \frac{sin ( 10 + 2 x )}{cos ( 10 + 2 x )} : \frac{- cos ( - 15 + 3 x ) cos ( 2 x + 10 ) + sin ( 10 + 2 x ) sin ( 3 x - 15 )}{sin ( 3 x - 15 ) cos ( 2 x + 10 )}

- \frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} + \frac{sin ( 10 + 2 x )}{cos ( 10 + 2 x )}

sin (- 15 + 3 x), cos (10 + 2 x)

的最小公倍数:

sin (3 x - 15) cos (2 x + 10)

sin (- 15 + 3 x), cos (10 + 2 x)

最小公倍数 (LCM)

计算出由出现在

sin (- 15 + 3 x)

或

cos (10 + 2 x)

中的因子组成的表达式

= sin (3 x - 15) cos (2 x + 10)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (3 x - 15) cos (2 x + 10)

对于

\frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} :

将分母和分子乘以

cos (2 x + 10)

\frac{cos ( - 15 + 3 x )}{sin ( - 15 + 3 x )} = \frac{cos ( - 15 + 3 x ) cos ( 2 x + 10 )}{sin ( - 15 + 3 x ) cos ( 2 x + 10 )}

对于

\frac{sin ( 10 + 2 x )}{cos ( 10 + 2 x )} :

将分母和分子乘以

sin (3 x - 15)

\frac{sin ( 10 + 2 x )}{cos ( 10 + 2 x )} = \frac{sin ( 10 + 2 x ) sin ( 3 x - 15 )}{cos ( 10 + 2 x ) sin ( 3 x - 15 )}

= - \frac{cos ( - 15 + 3 x ) cos ( 2 x + 10 )}{sin ( - 15 + 3 x ) cos ( 2 x + 10 )} + \frac{sin ( 10 + 2 x ) sin ( 3 x - 15 )}{cos ( 10 + 2 x ) sin ( 3 x - 15 )}

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{- cos ( - 15 + 3 x ) cos ( 2 x + 10 ) + sin ( 10 + 2 x ) sin ( 3 x - 15 )}{sin ( 3 x - 15 ) cos ( 2 x + 10 )}

= \frac{- cos ( - 15 + 3 x ) cos ( 2 x + 10 ) + sin ( 10 + 2 x ) sin ( 3 x - 15 )}{sin ( 3 x - 15 ) cos ( 2 x + 10 )}

\frac{- cos ( - 15 + 3 x ) cos ( 10 + 2 x ) + sin ( - 15 + 3 x ) sin ( 10 + 2 x )}{cos ( 10 + 2 x ) sin ( - 15 + 3 x )} = 0

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

- cos (- 15 + 3 x) cos (10 + 2 x) + sin (- 15 + 3 x) sin (10 + 2 x) = 0

使用三角恒等式改写

- cos (- 15 + 3 x) cos (10 + 2 x) + sin (- 15 + 3 x) sin (10 + 2 x)

使用角和恒等式:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

- cos (s) cos (t) + sin (s) sin (t) = - cos (s + t)

= - cos (- 15 + 3 x + 10 + 2 x)

- cos (- 15 + 3 x + 10 + 2 x) = 0

两边除以

- 1

- cos (- 15 + 3 x + 10 + 2 x) = 0

两边除以

- 1

\frac{- cos ( - 15 + 3 x + 10 + 2 x )}{- 1} = \frac{0}{- 1}

化简

cos (- 15 + 3 x + 10 + 2 x) = 0

cos (- 15 + 3 x + 10 + 2 x) = 0

cos (- 15 + 3 x + 10 + 2 x) = 0

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}

- 15 + 3 x + 10 + 2 x = \frac{π}{2} + 2 πn, - 15 + 3 x + 10 + 2 x = \frac{3 π}{2} + 2 πn

- 15 + 3 x + 10 + 2 x = \frac{π}{2} + 2 πn, - 15 + 3 x + 10 + 2 x = \frac{3 π}{2} + 2 πn

解

- 15 + 3 x + 10 + 2 x = \frac{π}{2} + 2 πn : x = \frac{π}{10} + \frac{2 πn}{5} + 1

- 15 + 3 x + 10 + 2 x = \frac{π}{2} + 2 πn

对同类项分组

3 x + 2 x - 15 + 10 = \frac{π}{2} + 2 πn

同类项相加:

3 x + 2 x = 5 x

5 x - 15 + 10 = \frac{π}{2} + 2 πn

数字相加/相减:

- 15 + 10 = - 5

5 x - 5 = \frac{π}{2} + 2 πn

将

5

到右边

5 x - 5 = \frac{π}{2} + 2 πn

两边加上

5

5 x - 5 + 5 = \frac{π}{2} + 2 πn + 5

化简

5 x = \frac{π}{2} + 2 πn + 5

5 x = \frac{π}{2} + 2 πn + 5

两边除以

5

5 x = \frac{π}{2} + 2 πn + 5

两边除以

5

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

化简

\frac{5 x}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + \frac{5}{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{π}{2}}{5} + \frac{2 πn}{5} + \frac{5}{5} : \frac{π}{10} + \frac{2 πn}{5} + 1

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

使用法则

\frac{a}{a} = 1

\frac{5}{5} = 1

= \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + 1

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

使用分式法则:

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

= \frac{π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{π}{10}

= \frac{π}{10} + \frac{2 πn}{5} + 1

x = \frac{π}{10} + \frac{2 πn}{5} + 1

x = \frac{π}{10} + \frac{2 πn}{5} + 1

x = \frac{π}{10} + \frac{2 πn}{5} + 1

解

- 15 + 3 x + 10 + 2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

- 15 + 3 x + 10 + 2 x = \frac{3 π}{2} + 2 πn

对同类项分组

3 x + 2 x - 15 + 10 = \frac{3 π}{2} + 2 πn

同类项相加:

3 x + 2 x = 5 x

5 x - 15 + 10 = \frac{3 π}{2} + 2 πn

数字相加/相减:

- 15 + 10 = - 5

5 x - 5 = \frac{3 π}{2} + 2 πn

将

5

到右边

5 x - 5 = \frac{3 π}{2} + 2 πn

两边加上

5

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

化简

5 x = \frac{3 π}{2} + 2 πn + 5

5 x = \frac{3 π}{2} + 2 πn + 5

两边除以

5

5 x = \frac{3 π}{2} + 2 πn + 5

两边除以

5

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

化简

\frac{5 x}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{5}{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{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{5}{5} : \frac{3 π}{10} + \frac{2 πn}{5} + 1

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{5}{5}

使用法则

\frac{a}{a} = 1

\frac{5}{5} = 1

= \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + 1

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

\frac{\frac{3 π}{2}}{5}

使用分式法则:

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

= \frac{3 π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{3 π}{10}

= \frac{3 π}{10} + \frac{2 πn}{5} + 1

x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

x = \frac{π}{10} + \frac{2 πn}{5} + 1, x = \frac{3 π}{10} + \frac{2 πn}{5} + 1

作图

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