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受欢迎的三角函数 >

tan(x)+sqrt(3)=sec(x)

解答

tan (x) + 3 = sec (x)

解答

x = \frac{11 π}{6} + 2 πn

+1

度数

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

求解步骤

tan (x) + 3 = sec (x)

两边减去

sec (x)

tan (x) + 3 - sec (x) = 0

用 sin, cos 表示

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

化简

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

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

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 )} + 3

将项转换为分式:

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

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

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

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

sin (x) - 1 + 3 cos (x) = 0

两边减去

3 cos (x)

sin (x) - 1 = - 3 cos (x)

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

化简

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

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

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

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - 1, b = sin (x)

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

化简

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

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

去除括号:

(- a) = - a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 \cdot sin (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin (x)

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

将解代入原方程进行验证

将它们代入

tan (x) + 3 = sec (x)

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:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

tan (x) + 3 = sec (x)

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

\infty = \infty

\Rightarrow 真

检验

\frac{7 π}{6} + 2 πn

的解:

假

\frac{7 π}{6} + 2 πn

代入

n = 1

\frac{7 π}{6} + 2 π 1

对于

tan (x) + 3 = sec (x)

代入

x = \frac{7 π}{6} + 2 π 1

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

整理后得

2.30940 \dots = - 1.15470 \dots

\Rightarrow 假

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) + 3 = sec (x)

代入

x = \frac{11 π}{6} + 2 π 1

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

整理后得

1.15470 \dots = 1.15470 \dots

\Rightarrow 真

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

因为方程对以下值无定义:

\frac{π}{2} + 2 πn

x = \frac{11 π}{6} + 2 πn

作图

流行的例子

16sec^2(θ)-1=0

16 sec^{2} (θ) - 1 = 0

cos(4y)=2cos(2y)-1

cos (4 y) = 2 cos (2 y) - 1

csc(x)cot(x)=2sqrt(3)

csc (x) cot (x) = 23

cos^2(θ)+2sin(θ)+1=0

cos^{2} (θ) + 2 sin (θ) + 1 = 0

arcsec(x)=arcsec(2)

arcsec (x) = arcsec (2)