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受欢迎的三角函数 >

cot(3θ)+1=csc(3θ)

解答

cot (3 θ) + 1 = csc (3 θ)

解答

θ = \frac{2 πn}{3} + \frac{π}{6}

+1

度数

θ = 3 0^{\circ} + 12 0^{\circ} n

求解步骤

cot (3 θ) + 1 = csc (3 θ)

两边减去

csc (3 θ)

cot (3 θ) + 1 - csc (3 θ) = 0

用 sin, cos 表示

1 + cot (3 θ) - csc (3 θ)

使用基本三角恒等式:

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

= 1 + \frac{cos ( 3 θ )}{sin ( 3 θ )} - csc (3 θ)

使用基本三角恒等式:

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

= 1 + \frac{cos ( 3 θ )}{sin ( 3 θ )} - \frac{1}{sin ( 3 θ )}

化简

1 + \frac{cos ( 3 θ )}{sin ( 3 θ )} - \frac{1}{sin ( 3 θ )} : \frac{sin ( 3 θ ) + cos ( 3 θ ) - 1}{sin ( 3 θ )}

1 + \frac{cos ( 3 θ )}{sin ( 3 θ )} - \frac{1}{sin ( 3 θ )}

合并分式

\frac{cos ( 3 θ )}{sin ( 3 θ )} - \frac{1}{sin ( 3 θ )} : \frac{cos ( 3 θ ) - 1}{sin ( 3 θ )}

使用法则

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

= \frac{cos ( 3 θ ) - 1}{sin ( 3 θ )}

= 1 + \frac{cos ( 3 θ ) - 1}{sin ( 3 θ )}

将项转换为分式:

1 = \frac{1 sin ( 3 θ )}{sin ( 3 θ )}

= \frac{1 \cdot sin ( 3 θ )}{sin ( 3 θ )} + \frac{cos ( 3 θ ) - 1}{sin ( 3 θ )}

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 sin ( 3 θ ) + cos ( 3 θ ) - 1}{sin ( 3 θ )}

乘以:

1 \cdot sin (3 θ) = sin (3 θ)

= \frac{sin ( 3 θ ) + cos ( 3 θ ) - 1}{sin ( 3 θ )}

= \frac{sin ( 3 θ ) + cos ( 3 θ ) - 1}{sin ( 3 θ )}

\frac{- 1 + cos ( 3 θ ) + sin ( 3 θ )}{sin ( 3 θ )} = 0

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

- 1 + cos (3 θ) + sin (3 θ) = 0

使用三角恒等式改写

- 1 + cos (3 θ) + sin (3 θ)

sin (3 θ) + cos (3 θ) = 2 sin (3 θ + \frac{π}{4})

sin (3 θ) + cos (3 θ)

改写为

= 2 (\frac{1}{2} sin (3 θ) + \frac{1}{2} cos (3 θ))

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (3 θ) + sin (\frac{π}{4}) cos (3 θ))

使用角和恒等式:

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

= 2 sin (3 θ + \frac{π}{4})

= - 1 + 2 sin (3 θ + \frac{π}{4})

- 1 + 2 sin (3 θ + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (3 θ + \frac{π}{4}) = 0

两边加上

1

- 1 + 2 sin (3 θ + \frac{π}{4}) + 1 = 0 + 1

化简

2 sin (3 θ + \frac{π}{4}) = 1

2 sin (3 θ + \frac{π}{4}) = 1

两边除以

2

2 sin (3 θ + \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( 3 θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( 3 θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( 3 θ + \frac{π}{4} )}{2} : sin (3 θ + \frac{π}{4})

\frac{2 sin ( 3 θ + \frac{π}{4} )}{2}

约分:

2

= sin (3 θ + \frac{π}{4})

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (3 θ + \frac{π}{4}) = \frac{2}{2}

sin (3 θ + \frac{π}{4}) = \frac{2}{2}

sin (3 θ + \frac{π}{4}) = \frac{2}{2}

sin (3 θ + \frac{π}{4}) = \frac{2}{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}

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

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

解

3 θ + \frac{π}{4} = \frac{π}{4} + 2 πn : θ = \frac{2 πn}{3}

3 θ + \frac{π}{4} = \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

3 θ + \frac{π}{4} - \frac{π}{4} = \frac{π}{4} + 2 πn - \frac{π}{4}

化简

3 θ = 2 πn

两边除以

3

3 θ = 2 πn

两边除以

3

\frac{3 θ}{3} = \frac{2 πn}{3}

化简

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

解

3 θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn : θ = \frac{2 πn}{3} + \frac{π}{6}

3 θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

3 θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

3 θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

3 θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

3 θ + \frac{π}{4} - \frac{π}{4} : 3 θ

3 θ + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= 3 θ

化简

\frac{3 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{2}

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

3 θ = 2 πn + \frac{π}{2}

3 θ = 2 πn + \frac{π}{2}

3 θ = 2 πn + \frac{π}{2}

两边除以

3

3 θ = 2 πn + \frac{π}{2}

两边除以

3

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

化简

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

化简

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

数字相除:

\frac{3}{3} = 1

= θ

化简

\frac{2 πn}{3} + \frac{\frac{π}{2}}{3} : \frac{2 πn}{3} + \frac{π}{6}

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

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

使用分式法则:

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

= \frac{π}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{2 πn}{3} + \frac{π}{6}

θ = \frac{2 πn}{3} + \frac{π}{6}

θ = \frac{2 πn}{3} + \frac{π}{6}

θ = \frac{2 πn}{3} + \frac{π}{6}

θ = \frac{2 πn}{3}, θ = \frac{2 πn}{3} + \frac{π}{6}

因为方程对以下值无定义:

\frac{2 πn}{3}

θ = \frac{2 πn}{3} + \frac{π}{6}

作图

流行的例子

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

sin (θ) = \frac{5}{11}

4 = - 8 cos (θ)

2cos^2(θ)-5cos(θ)=3

2 cos^{2} (θ) - 5 cos (θ) = 3

12sin^2(x)-20sin(x)=-8

12 sin^{2} (x) - 20 sin (x) = - 8