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受欢迎的三角函数 >

(sec(x)+tan(x))/(cot(x)+cos(x))=sec^2(x)

解答

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} = sec^{2} (x)

解答

x \in R 无解

求解步骤

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} = sec^{2} (x)

两边减去

sec^{2} (x)

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} - sec^{2} (x) = 0

化简

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} - sec^{2} (x) : \frac{sec ( x ) + tan ( x ) - sec ^{2} ( x ) ( cot ( x ) + cos ( x ) )}{cot ( x ) + cos ( x )}

\frac{sec ( x ) + tan ( x )}{cot ( x ) + cos ( x )} - sec^{2} (x)

将项转换为分式:

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

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

\frac{sec ( x ) + tan ( x ) - sec ^{2} ( x ) ( cot ( x ) + cos ( x ) )}{cot ( x ) + cos ( x )} = 0

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

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

合并分式

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

使用法则

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

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

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

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

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

化简

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

cos (x) + \frac{cos ( x )}{sin ( x )}

将项转换为分式:

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

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

分式相乘:

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

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

(cos (x) sin (x) + cos (x)) \cdot 1 = cos (x) sin (x) + cos (x)

(cos (x) sin (x) + cos (x)) \cdot 1

乘以:

(cos (x) sin (x) + cos (x)) \cdot 1 = (cos (x) sin (x) + cos (x))

= (cos (x) sin (x) + cos (x))

去除括号:

(a) = a

= cos (x) sin (x) + cos (x)

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

因式分解出通项

cos (x)

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

约分:

cos (x)

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

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

cos (x), sin (x) cos (x)

的最小公倍数:

cos (x) sin (x)

cos (x), sin (x) cos (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

sin (x) cos (x)

中的因子组成的表达式

= cos (x) sin (x)

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

cos (x) sin (x)

对于

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

将分母和分子乘以

sin (x)

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

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

乘开

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

(1 + sin (x)) sin (x) - (sin (x) + 1)

= sin (x) (1 + sin (x)) - (sin (x) + 1)

乘开

sin (x) (1 + sin (x)) : sin (x) + sin^{2} (x)

sin (x) (1 + sin (x))

使用分配律:

a (b + c) = ab + a c

a = sin (x), b = 1, c = sin (x)

= sin (x) \cdot 1 + sin (x) sin (x)

= 1 \cdot sin (x) + sin (x) sin (x)

化简

1 \cdot sin (x) + sin (x) sin (x) : sin (x) + sin^{2} (x)

1 \cdot sin (x) + sin (x) sin (x)

1 \cdot sin (x) = sin (x)

1 \cdot sin (x)

乘以:

1 \cdot sin (x) = sin (x)

= sin (x)

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

= sin (x) + sin^{2} (x)

= sin (x) + sin^{2} (x)

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

- (sin (x) + 1) : - sin (x) - 1

- (sin (x) + 1)

打开括号

= - (sin (x)) - (1)

使用加减运算法则

+ (- a) = - a

= - sin (x) - 1

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

同类项相加:

sin (x) - sin (x) = 0

= sin^{2} (x) - 1

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

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

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

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

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

用替代法求解

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

令

: sin (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 = sin (x)

代回

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

sin (x) = - 1

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

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

因为方程对以下值无定义:

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

x \in R 无解

作图

流行的例子

cos (x) = csc (x)

1+cos(2t)=cos^2(t)

1 + cos (2 t) = cos^{2} (t)

2sin(2x+(3pi)/2)+1=0

2 sin (2 x + \frac{3 π}{2}) + 1 = 0

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

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

2cos^2(θ)-2sin^2(θ)=sqrt(2)

2 cos^{2} (θ) - 2 sin^{2} (θ) = 2