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受欢迎的三角函数 >

tan^4(x)-2=tan^2(x)+sec^2(x)

解答

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)

解答

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

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

求解步骤

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)

两边减去

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

tan^{4} (x) - 2 - tan^{2} (x) - sec^{2} (x) = 0

使用三角恒等式改写

- 2 - sec^{2} (x) - tan^{2} (x) + tan^{4} (x)

使用毕达哥拉斯恒等式:

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

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

化简

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

- 2 - (tan^{2} (x) + 1) - tan^{2} (x) + tan^{4} (x)

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

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

化简

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

- 2 - tan^{2} (x) - 1 - tan^{2} (x) + tan^{4} (x)

对同类项分组

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

同类项相加:

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

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

数字相减:

- 2 - 1 = - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

= tan^{4} (x) - 2 tan^{2} (x) - 3

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

用替代法求解

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

令

: tan (x) = u

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

- 3 + u^{4} - 2 u^{2} = 0 : u = 3, u = - 3, u = i, u = - i

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

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 2 u^{2} - 3 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 2 v - 3 = 0

解

v^{2} - 2 v - 3 = 0 : v = 3, v = - 1

v^{2} - 2 v - 3 = 0

使用求根公式求解

v^{2} - 2 v - 3 = 0

二次方程求根公式:

若

a = 1, b = - 2, c = - 3

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

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

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数字相加:

4 + 12 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

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

将解分隔开

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

v = \frac{- ( - 2 ) + 4}{2 \cdot 1} : 3

\frac{- ( - 2 ) + 4}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 + 4}{2 \cdot 1}

数字相加:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

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

\frac{- ( - 2 ) - 4}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 - 4}{2 \cdot 1}

数字相减:

2 - 4 = - 2

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

数字相乘:

2 \cdot 1 = 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:

v = 3, v = - 1

v = 3, v = - 1

代回

v = u^{2}

，求解

u

解

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

解

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

对于

x^{2} = f (a)

解为

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

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

u = 3, u = - 3, u = i, u = - i

u = tan (x)

代回

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

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

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) = - 3 : x = \frac{2 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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

tan (x) = i :

无解

tan (x) = i

无解

tan (x) = - i :

无解

tan (x) = - i

无解

合并所有解

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

作图

流行的例子

cos(x-pi/4)=-sin(x)

cos (x - \frac{π}{4}) = - sin (x)

sin(x)= 5/13 ,cos(x)

sin (x) = \frac{5}{13}, cos (x)

tan(θ)=(11.9)/(10)

tan (θ) = \frac{11.9}{10}

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

csc(θ)+2.402=0

csc (θ) + 2.402 = 0