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受欢迎的三角函数 >

(tan(x)-sec(x))^2=3

解答

(tan (x) - sec (x))^{2} = 3

解答

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

(tan (x) - sec (x))^{2} = 3

两边减去

3

(tan (x) - sec (x))^{2} - 3 = 0

用 sin, cos 表示

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3 = 0

化简

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3 : \frac{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )})^{2} - 3

合并分式

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

使用法则

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

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

= (\frac{sin ( x ) - 1}{cos ( x )})^{2} - 3

使用指数法则:

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

= \frac{( sin ( x ) - 1 ) ^{2}}{cos ^{2} ( x )} - 3

将项转换为分式:

3 = \frac{3 cos ^{2} ( x )}{cos ^{2} ( x )}

= \frac{( sin ( x ) - 1 ) ^{2}}{cos ^{2} ( x )} - \frac{3 cos ^{2} ( x )}{cos ^{2} ( 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{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )}

\frac{( sin ( x ) - 1 ) ^{2} - 3 cos ^{2} ( x )}{cos ^{2} ( x )} = 0

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

(sin (x) - 1)^{2} - 3 cos^{2} (x) = 0

两边加上

3 cos^{2} (x)

sin^{2} (x) - 2 sin (x) + 1 = 3 cos^{2} (x)

两边进行平方

(sin^{2} (x) - 2 sin (x) + 1)^{2} = (3 cos^{2} (x))^{2}

两边减去

(3 cos^{2} (x))^{2}

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) = 0

分解

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x) : (sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x)) (sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x))

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

将

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

改写为

(sin^{2} (x) - 2 sin (x) + 1)^{2} - (3 cos^{2} (x))^{2}

(sin^{2} (x) - 2 sin (x) + 1)^{2} - 9 cos^{4} (x)

将

9

改写为

3^{2}

= (sin^{2} (x) - 2 sin (x) + 1)^{2} - 3^{2} cos^{4} (x)

使用指数法则:

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

cos^{4} (x) = (cos^{2} (x))^{2}

= (sin^{2} (x) - 2 sin (x) + 1)^{2} - 3^{2} (cos^{2} (x))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (x))^{2} = (3 cos^{2} (x))^{2}

= (sin^{2} (x) - 2 sin (x) + 1)^{2} - (3 cos^{2} (x))^{2}

= (sin^{2} (x) - 2 sin (x) + 1)^{2} - (3 cos^{2} (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(sin^{2} (x) - 2 sin (x) + 1)^{2} - (3 cos^{2} (x))^{2} = ((sin^{2} (x) - 2 sin (x) + 1) + 3 cos^{2} (x)) ((sin^{2} (x) - 2 sin (x) + 1) - 3 cos^{2} (x))

= ((sin^{2} (x) - 2 sin (x) + 1) + 3 cos^{2} (x)) ((sin^{2} (x) - 2 sin (x) + 1) - 3 cos^{2} (x))

整理后得

= (sin^{2} (x) + 3 cos^{2} (x) - 2 sin (x) + 1) (sin^{2} (x) - 3 cos^{2} (x) - 2 sin (x) + 1)

(sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x)) (sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x)) = 0

分别求解每个部分

sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x) = 0 or sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x) = 0

sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn

sin^{2} (x) - 2 sin (x) + 1 + 3 cos^{2} (x) = 0

使用三角恒等式改写

1 + sin^{2} (x) - 2 sin (x) + 3 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= 1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x))

化简

1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x)) : - 2 sin^{2} (x) - 2 sin (x) + 4

1 + sin^{2} (x) - 2 sin (x) + 3 (1 - sin^{2} (x))

乘开

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (x)

= 1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x)

化简

1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x) : - 2 sin^{2} (x) - 2 sin (x) + 4

1 + sin^{2} (x) - 2 sin (x) + 3 - 3 sin^{2} (x)

对同类项分组

= sin^{2} (x) - 2 sin (x) - 3 sin^{2} (x) + 1 + 3

同类项相加:

sin^{2} (x) - 3 sin^{2} (x) = - 2 sin^{2} (x)

= - 2 sin^{2} (x) - 2 sin (x) + 1 + 3

数字相加:

1 + 3 = 4

= - 2 sin^{2} (x) - 2 sin (x) + 4

= - 2 sin^{2} (x) - 2 sin (x) + 4

= - 2 sin^{2} (x) - 2 sin (x) + 4

4 - 2 sin (x) - 2 sin^{2} (x) = 0

用替代法求解

4 - 2 sin (x) - 2 sin^{2} (x) = 0

令

: sin (x) = u

4 - 2 u - 2 u^{2} = 0

4 - 2 u - 2 u^{2} = 0 : u = - 2, u = 1

4 - 2 u - 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 u^{2} - 2 u + 4 = 0

使用求根公式求解

- 2 u^{2} - 2 u + 4 = 0

二次方程求根公式:

若

a = - 2, b = - 2, c = 4

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 4}{2 ( - 2 )}

(- 2)^{2} - 4 (- 2) \cdot 4 = 6

(- 2)^{2} - 4 (- 2) \cdot 4

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 4

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 2 \cdot 4

数字相乘:

4 \cdot 2 \cdot 4 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 6}{2 ( - 2 )}, u_{2} = \frac{- ( - 2 ) - 6}{2 ( - 2 )}

u = \frac{- ( - 2 ) + 6}{2 ( - 2 )} : - 2

\frac{- ( - 2 ) + 6}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 + 6}{- 2 \cdot 2}

数字相加:

2 + 6 = 8

= \frac{8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 2 ) - 6}{2 ( - 2 )} : 1

\frac{- ( - 2 ) - 6}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 - 6}{- 2 \cdot 2}

数字相减:

2 - 6 = - 4

= \frac{- 4}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - 2, u = 1

u = sin (x)

代回

sin (x) = - 2, sin (x) = 1

sin (x) = - 2, sin (x) = 1

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

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

合并所有解

x = \frac{π}{2} + 2 πn

sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin^{2} (x) - 2 sin (x) + 1 - 3 cos^{2} (x) = 0

使用三角恒等式改写

1 + sin^{2} (x) - 2 sin (x) - 3 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= 1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x))

化简

1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x)) : 4 sin^{2} (x) - 2 sin (x) - 2

1 + sin^{2} (x) - 2 sin (x) - 3 (1 - sin^{2} (x))

乘开

- 3 (1 - sin^{2} (x)) : - 3 + 3 sin^{2} (x)

- 3 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (x)

= - 3 \cdot 1 - (- 3) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (x)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (x)

= 1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x)

化简

1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x) : 4 sin^{2} (x) - 2 sin (x) - 2

1 + sin^{2} (x) - 2 sin (x) - 3 + 3 sin^{2} (x)

对同类项分组

= sin^{2} (x) - 2 sin (x) + 3 sin^{2} (x) + 1 - 3

同类项相加:

sin^{2} (x) + 3 sin^{2} (x) = 4 sin^{2} (x)

= 4 sin^{2} (x) - 2 sin (x) + 1 - 3

数字相加/相减:

1 - 3 = - 2

= 4 sin^{2} (x) - 2 sin (x) - 2

= 4 sin^{2} (x) - 2 sin (x) - 2

= 4 sin^{2} (x) - 2 sin (x) - 2

- 2 - 2 sin (x) + 4 sin^{2} (x) = 0

用替代法求解

- 2 - 2 sin (x) + 4 sin^{2} (x) = 0

令

: sin (x) = u

- 2 - 2 u + 4 u^{2} = 0

- 2 - 2 u + 4 u^{2} = 0 : u = 1, u = - \frac{1}{2}

- 2 - 2 u + 4 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

4 u^{2} - 2 u - 2 = 0

使用求根公式求解

4 u^{2} - 2 u - 2 = 0

二次方程求根公式:

若

a = 4, b = - 2, c = - 2

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 2) = 6

(- 2)^{2} - 4 \cdot 4 (- 2)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 4 \cdot 2

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 4}

u = \frac{- ( - 2 ) + 6}{2 \cdot 4} : 1

\frac{- ( - 2 ) + 6}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 + 6}{2 \cdot 4}

数字相加:

2 + 6 = 8

= \frac{8}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{8}{8}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 6}{2 \cdot 4} : - \frac{1}{2}

\frac{- ( - 2 ) - 6}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 - 6}{2 \cdot 4}

数字相减:

2 - 6 = - 4

= \frac{- 4}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{8}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

二次方程组的解是:

u = 1, u = - \frac{1}{2}

u = sin (x)

代回

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

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

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) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

将解代入原方程进行验证

将它们代入

(tan (x) - sec (x))^{2} = 3

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

假

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

(tan (x) - sec (x))^{2} = 3

代入

x = \frac{π}{2} + 2 π 1

(tan (\frac{π}{2} + 2 π 1) - sec (\frac{π}{2} + 2 π 1))^{2} = 3

未定义

\Rightarrow 假

检验

\frac{7 π}{6} + 2 πn

的解:

真

\frac{7 π}{6} + 2 πn

代入

n = 1

\frac{7 π}{6} + 2 π 1

对于

(tan (x) - sec (x))^{2} = 3

代入

x = \frac{7 π}{6} + 2 π 1

(tan (\frac{7 π}{6} + 2 π 1) - sec (\frac{7 π}{6} + 2 π 1))^{2} = 3

整理后得

3 = 3

\Rightarrow 真

检验

\frac{11 π}{6} + 2 πn

的解:

真

\frac{11 π}{6} + 2 πn

代入

n = 1

\frac{11 π}{6} + 2 π 1

对于

(tan (x) - sec (x))^{2} = 3

代入

x = \frac{11 π}{6} + 2 π 1

(tan (\frac{11 π}{6} + 2 π 1) - sec (\frac{11 π}{6} + 2 π 1))^{2} = 3

整理后得

3 = 3

\Rightarrow 真

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

作图

流行的例子

solvefor y,ln(x^2+10)+csc(y)=c

so l v e f or y, ln (x^{2} + 10) + csc (y) = c

sin (a) = \frac{7}{15}

(3sqrt(3))/(14)=sin(x)

\frac{3 3}{14} = sin (x)

6cos(x)=2+2cos(x)

6 cos (x) = 2 + 2 cos (x)

cosh(z)=1,cosh(z)=-2

cosh (z) = 1, cosh (z) = - 2