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受欢迎的三角函数 >

cos(x)sec(x)cot^2(x)=csc^2(x)

解答

cos (x) \cdot sec (x) \cdot cot^{2} (x) = csc^{2} (x)

解答

x \in R 无解

求解步骤

cos (x) sec (x) cot^{2} (x) = csc^{2} (x)

两边减去

csc^{2} (x)

cot^{2} (x) cos (x) sec (x) - csc^{2} (x) = 0

用 sin, cos 表示

- csc^{2} (x) + cos (x) cot^{2} (x) sec (x)

使用基本三角恒等式:

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

= - (\frac{1}{sin ( x )})^{2} + cos (x) cot^{2} (x) sec (x)

使用基本三角恒等式:

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

= - (\frac{1}{sin ( x )})^{2} + cos (x) (\frac{cos ( x )}{sin ( x )})^{2} sec (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

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

化简

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

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

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

cos (x) (\frac{cos ( x )}{sin ( x )})^{2} \frac{1}{cos ( x )} = \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

cos (x) (\frac{cos ( x )}{sin ( x )})^{2} \frac{1}{cos ( x )}

分式相乘:

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

= \frac{1 \cdot cos ( x ) ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}{cos ( x )}

约分:

cos (x)

= 1 \cdot (\frac{cos ( x )}{sin ( x )})^{2}

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

(\frac{cos ( x )}{sin ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= 1 \cdot \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

乘以:

1 \cdot \frac{cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

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

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

使用法则

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

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

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

\frac{- 1 + cos ^{2} ( x )}{sin ^{2} ( x )} = 0

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

- 1 + cos^{2} (x) = 0

用替代法求解

- 1 + cos^{2} (x) = 0

令

: cos (x) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

将

1

到右边

- 1 + u^{2} = 0

两边加上

1

- 1 + u^{2} + 1 = 0 + 1

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = cos (x)

代回

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn

因为方程对以下值无定义:

2 πn, π + 2 πn

x \in R 无解

作图

流行的例子

4cot(x)=sqrt(3)csc(x)

4 cot (x) = 3 csc (x)

sin (θ) = \frac{7}{15}

cos(x)=(sqrt(5))/3

cos (x) = \frac{5}{3}

cos (x) = - \frac{7}{8}

sin(x)=-cos(2x)

sin (x) = - cos (2 x)