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受欢迎的三角函数 >

tan(3x)+cos(6x)=1

解答

tan (3 x) + cos (6 x) = 1

解答

x = \frac{2 πn}{3}, x = \frac{π + 2 πn}{3}, x = \frac{π + 4 πn}{12}

+1

度数

x = 0^{\circ} + 12 0^{\circ} n, x = 6 0^{\circ} + 12 0^{\circ} n, x = 1 5^{\circ} + 6 0^{\circ} n

求解步骤

tan (3 x) + cos (6 x) = 1

两边减去

1

tan (3 x) + cos (6 x) - 1 = 0

令

: u = 3 x

tan (u) + cos (2 u) - 1 = 0

用 sin, cos 表示

- 1 + cos (2 u) + tan (u)

使用基本三角恒等式:

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

= - 1 + cos (2 u) + \frac{sin ( u )}{cos ( u )}

化简

- 1 + cos (2 u) + \frac{sin ( u )}{cos ( u )} : \frac{- cos ( u ) + cos ( 2 u ) cos ( u ) + sin ( u )}{cos ( u )}

- 1 + cos (2 u) + \frac{sin ( u )}{cos ( u )}

将项转换为分式:

1 = \frac{1 cos ( u )}{cos ( u )}, cos (2 u) = \frac{cos ( 2 u ) cos ( u )}{cos ( u )}

= - \frac{1 \cdot cos ( u )}{cos ( u )} + \frac{cos ( 2 u ) cos ( u )}{cos ( u )} + \frac{sin ( u )}{cos ( u )}

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{- 1 \cdot cos ( u ) + cos ( 2 u ) cos ( u ) + sin ( u )}{cos ( u )}

乘以:

1 \cdot cos (u) = cos (u)

= \frac{- cos ( u ) + cos ( 2 u ) cos ( u ) + sin ( u )}{cos ( u )}

= \frac{- cos ( u ) + cos ( 2 u ) cos ( u ) + sin ( u )}{cos ( u )}

\frac{- cos ( u ) + sin ( u ) + cos ( 2 u ) cos ( u )}{cos ( u )} = 0

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

- cos (u) + sin (u) + cos (2 u) cos (u) = 0

使用三角恒等式改写

- cos (u) + sin (u) + cos (2 u) cos (u)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= - cos (u) + sin (u) + (1 - 2 sin^{2} (u)) cos (u)

化简

- cos (u) + sin (u) + (1 - 2 sin^{2} (u)) cos (u) : sin (u) - 2 sin^{2} (u) cos (u)

- cos (u) + sin (u) + (1 - 2 sin^{2} (u)) cos (u)

= - cos (u) + sin (u) + cos (u) (1 - 2 sin^{2} (u))

乘开

cos (u) (1 - 2 sin^{2} (u)) : cos (u) - 2 sin^{2} (u) cos (u)

cos (u) (1 - 2 sin^{2} (u))

使用分配律:

a (b - c) = ab - a c

a = cos (u), b = 1, c = 2 sin^{2} (u)

= cos (u) \cdot 1 - cos (u) \cdot 2 sin^{2} (u)

= 1 \cdot cos (u) - 2 sin^{2} (u) cos (u)

乘以:

1 \cdot cos (u) = cos (u)

= cos (u) - 2 sin^{2} (u) cos (u)

= - cos (u) + sin (u) + cos (u) - 2 sin^{2} (u) cos (u)

同类项相加:

- cos (u) + cos (u) = 0

= sin (u) - 2 sin^{2} (u) cos (u)

= sin (u) - 2 sin^{2} (u) cos (u)

sin (u) - 2 cos (u) sin^{2} (u) = 0

分解

sin (u) - 2 cos (u) sin^{2} (u) : sin (u) (1 - 2 sin (u) cos (u))

sin (u) - 2 cos (u) sin^{2} (u)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{2} (u) cos (u) = sin (u) sin (u)

= sin (u) - 2 sin (u) sin (u)

因式分解出通项

sin (u)

= sin (u) (1 - 2 sin (u) cos (u))

sin (u) (1 - 2 sin (u) cos (u)) = 0

分别求解每个部分

sin (u) = 0 or 1 - 2 sin (u) cos (u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

1 - 2 sin (u) cos (u) = 0 : u = \frac{π}{4} + πn

1 - 2 sin (u) cos (u) = 0

使用三角恒等式改写

1 - 2 sin (u) cos (u)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

= 1 - sin (2 u)

1 - sin (2 u) = 0

将

1

到右边

1 - sin (2 u) = 0

两边减去

1

1 - sin (2 u) - 1 = 0 - 1

化简

- sin (2 u) = - 1

- sin (2 u) = - 1

两边除以

- 1

- sin (2 u) = - 1

两边除以

- 1

\frac{- sin ( 2 u )}{- 1} = \frac{- 1}{- 1}

化简

sin (2 u) = 1

sin (2 u) = 1

sin (2 u) = 1

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 u = \frac{π}{2} + 2 πn

2 u = \frac{π}{2} + 2 πn

解

2 u = \frac{π}{2} + 2 πn : u = \frac{π}{4} + πn

2 u = \frac{π}{2} + 2 πn

两边除以

2

2 u = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

数字相除:

\frac{2}{2} = 1

= u

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

u = \frac{π}{4} + πn

u = \frac{π}{4} + πn

u = \frac{π}{4} + πn

u = \frac{π}{4} + πn

合并所有解

u = 2 πn, u = π + 2 πn, u = \frac{π}{4} + πn

u = 3 x

代回

3 x = 2 πn : x = \frac{2 πn}{3}

3 x = 2 πn

两边除以

3

3 x = 2 πn

两边除以

3

\frac{3 x}{3} = \frac{2 πn}{3}

化简

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

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

3 x = π + 2 πn

两边除以

3

3 x = π + 2 πn

两边除以

3

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

化简

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

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

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

\frac{π}{3} + \frac{2 πn}{3}

使用法则

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

= \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

3 x = \frac{π}{4} + πn : x = \frac{π + 4 πn}{12}

3 x = \frac{π}{4} + πn

两边除以

3

3 x = \frac{π}{4} + πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{π}{4}}{3} + \frac{πn}{3} : \frac{π + 4 πn}{12}

\frac{\frac{π}{4}}{3} + \frac{πn}{3}

使用法则

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

= \frac{\frac{π}{4} + πn}{3}

化简

\frac{π}{4} + πn : \frac{π + 4 πn}{4}

\frac{π}{4} + πn

将项转换为分式:

πn = \frac{πn 4}{4}

= \frac{π}{4} + \frac{πn \cdot 4}{4}

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{π + πn \cdot 4}{4}

= \frac{\frac{π + 4 πn}{4}}{3}

使用分式法则:

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

= \frac{π + πn \cdot 4}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= \frac{π + 4 πn}{12}

x = \frac{π + 4 πn}{12}

x = \frac{π + 4 πn}{12}

x = \frac{π + 4 πn}{12}

x = \frac{2 πn}{3}, x = \frac{π + 2 πn}{3}, x = \frac{π + 4 πn}{12}

作图

流行的例子

sin^2(θ)=2cos(θ)+2,0<= θ<= 2pi

sin^{2} (θ) = 2 cos (θ) + 2, 0 \leq θ \leq 2 π

sin (x) = - 0.3

-sin(x)+cos(x)=0,-pi<= x<= pi

- sin (x) + cos (x) = 0, - π \leq x \leq π

sin^2(x)=2-2cos(x),0<= x<= 2pi

sin^{2} (x) = 2 - 2 cos (x), 0 \leq x \leq 2 π

tan(θ)=(sqrt(7))/(21)

tan (θ) = \frac{7}{21}