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受欢迎的三角函数 >

sec^2(x)-1+tan(x)-sqrt(3)tan(x)=sqrt(3)

解答

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

解答

x = \frac{π}{3} + πn, x = \frac{3 π}{4} + πn

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

3

sec^{2} (x) - 1 + tan (x) - 3 tan (x) - 3 = 0

使用三角恒等式改写

- 1 + sec^{2} (x) - 3 + tan (x) - 3 tan (x)

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

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

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

用替代法求解

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

令

: tan (x) = u

- 3 + u + u^{2} - 3 u = 0

- 3 + u + u^{2} - 3 u = 0 : u = 3, u = - 1

- 3 + u + u^{2} - 3 u = 0

改写成标准形式

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

u^{2} + (1 - 3) u - 3 = 0

使用求根公式求解

u^{2} + (1 - 3) u - 3 = 0

二次方程求根公式:

若

a = 1, b = 1 - 3, c = - 3

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 1 - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 1 - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(1 - 3)^{2} - 4 \cdot 1 \cdot (- 3) = 3 + 1

(1 - 3)^{2} - 4 \cdot 1 \cdot (- 3)

使用法则

- (- a) = a

= (1 - 3)^{2} + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 = 4

= (1 - 3)^{2} + 43

乘开

(1 - 3)^{2} + 43 : 4 + 23

(1 - 3)^{2} + 43

(1 - 3)^{2} : 4 - 23

使用完全平方公式:

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

a = 1, b = 3

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

化简

1^{2} - 2 \cdot 1 \cdot 3 + (3)^{2} : 4 - 23

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 3 = 23

2 \cdot 1 \cdot 3

数字相乘:

2 \cdot 1 = 2

= 23

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 3

= 1 - 23 + 3

数字相加:

1 + 3 = 4

= 4 - 23

= 4 - 23

= 4 - 23 + 43

同类项相加:

- 23 + 43 = 23

= 4 + 23

= 4 + 23

= 3 + 23 + 1

= (3)^{2} + 23 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (3)^{2} + 23 + 1^{2}

23 \cdot 1 = 23

23 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 23

= (3)^{2} + 23 \cdot 1 + 1^{2}

使用完全平方公式:

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

(3)^{2} + 23 \cdot 1 + 1^{2} = (3 + 1)^{2}

= (3 + 1)^{2}

使用根式运算法则:

n a^{n} = a

(3 + 1)^{2} = 3 + 1

= 3 + 1

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 3 + 1 )}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( 1 - 3 ) + 3 + 1}{2 \cdot 1}, u_{2} = \frac{- ( 1 - 3 ) - ( 3 + 1 )}{2 \cdot 1}

u = \frac{- ( 1 - 3 ) + 3 + 1}{2 \cdot 1} : 3

\frac{- ( 1 - 3 ) + 3 + 1}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- ( 1 - 3 ) + 3 + 1}{2}

乘开

- (1 - 3) + 3 + 1 : 23

- (1 - 3) + 3 + 1

- (1 - 3) : - 1 + 3

- (1 - 3)

打开括号

= - (1) - (- 3)

使用加减运算法则

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

= - 1 + 3

= - 1 + 3 + 3 + 1

化简

- 1 + 3 + 3 + 1 : 23

- 1 + 3 + 3 + 1

同类项相加:

3 + 3 = 23

= - 1 + 23 + 1

- 1 + 1 = 0

= 23

= 23

= \frac{2 3}{2}

数字相除:

\frac{2}{2} = 1

= 3

u = \frac{- ( 1 - 3 ) - ( 3 + 1 )}{2 \cdot 1} : - 1

\frac{- ( 1 - 3 ) - ( 3 + 1 )}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- ( 1 - 3 ) - ( 1 + 3 )}{2}

乘开

- (1 - 3) - (3 + 1) : - 2

- (1 - 3) - (3 + 1)

- (1 - 3) : - 1 + 3

- (1 - 3)

打开括号

= - (1) - (- 3)

使用加减运算法则

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

= - 1 + 3

= - 1 + 3 - (3 + 1)

- (3 + 1) : - 3 - 1

- (3 + 1)

打开括号

= - (3) - (1)

使用加减运算法则

+ (- a) = - a

= - 3 - 1

= - 1 + 3 - 3 - 1

化简

- 1 + 3 - 3 - 1 : - 2

- 1 + 3 - 3 - 1

同类项相加:

3 - 3 = 0

= - 1 - 1

数字相减:

- 1 - 1 = - 2

= - 2

= - 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 3, u = - 1

u = tan (x)

代回

tan (x) = 3, tan (x) = - 1

tan (x) = 3, tan (x) = - 1

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

tan (x) = 3

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

tan (x) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = \frac{π}{3} + πn, x = \frac{3 π}{4} + πn

作图

流行的例子

sqrt(3)sec(2x)-2=0

3 sec (2 x) - 2 = 0

2sin^2(θ)+3sin(θ)=0

2 sin^{2} (θ) + 3 sin (θ) = 0

sin(x+pi/6)+sin(x-pi/6)=-(sqrt(3))/2

sin (x + \frac{π}{6}) + sin (x - \frac{π}{6}) = - \frac{3}{2}

sqrt(3)csc((2x)/3)-2=0

3 csc (\frac{2 x}{3}) - 2 = 0

cos (θ) = \frac{8}{10}