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受欢迎的三角函数 >

tan^2(x)-4tan(x)+1=0

解答

tan^{2} (x) - 4 tan (x) + 1 = 0

解答

x = 1.30899 \dots + πn, x = 0.26179 \dots + πn

+1

度数

x = 7 5^{\circ} + 18 0^{\circ} n, x = 1 5^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (x) - 4 tan (x) + 1 = 0

用替代法求解

tan^{2} (x) - 4 tan (x) + 1 = 0

令

: tan (x) = u

u^{2} - 4 u + 1 = 0

u^{2} - 4 u + 1 = 0 : u = 2 + 3, u = 2 - 3

u^{2} - 4 u + 1 = 0

使用求根公式求解

u^{2} - 4 u + 1 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

数字相减:

16 - 4 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{4 + 2 3}{2}

分解

4 + 23 : 2 (2 + 3)

4 + 23

改写为

= 2 \cdot 2 + 23

因式分解出通项

2

= 2 (2 + 3)

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

数字相除:

\frac{2}{2} = 1

= 2 + 3

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

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

使用法则

- (- a) = a

= \frac{4 - 2 3}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 - 2 3}{2}

分解

4 - 23 : 2 (2 - 3)

4 - 23

改写为

= 2 \cdot 2 - 23

因式分解出通项

2

= 2 (2 - 3)

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

数字相除:

\frac{2}{2} = 1

= 2 - 3

二次方程组的解是:

u = 2 + 3, u = 2 - 3

u = tan (x)

代回

tan (x) = 2 + 3, tan (x) = 2 - 3

tan (x) = 2 + 3, tan (x) = 2 - 3

tan (x) = 2 + 3 : x = arctan (2 + 3) + πn

tan (x) = 2 + 3

使用反三角函数性质

tan (x) = 2 + 3

tan (x) = 2 + 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (2 + 3) + πn

x = arctan (2 + 3) + πn

tan (x) = 2 - 3 : x = arctan (2 - 3) + πn

tan (x) = 2 - 3

使用反三角函数性质

tan (x) = 2 - 3

tan (x) = 2 - 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (2 - 3) + πn

x = arctan (2 - 3) + πn

合并所有解

x = arctan (2 + 3) + πn, x = arctan (2 - 3) + πn

以小数形式表示解

x = 1.30899 \dots + πn, x = 0.26179 \dots + πn

作图

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