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受欢迎的三角函数 >

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

解答

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

解答

x = 1.17809 \dots + πn, x = - 0.39269 \dots + πn

+1

度数

x = 67. 5^{\circ} + 18 0^{\circ} n, x = - 22. 5^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: tan (x) = u

u^{2} - 2 u = 1

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

u^{2} - 2 u = 1

将

1

para o lado esquerdo

u^{2} - 2 u = 1

两边减去

1

u^{2} - 2 u - 1 = 1 - 1

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

使用法则

- (- a) = a

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

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 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

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

数字相除:

\frac{2}{2} = 1

= 1 + 2

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

分解

2 - 22 : 2 (1 - 2)

2 - 22

改写为

= 2 \cdot 1 - 22

因式分解出通项

2

= 2 (1 - 2)

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

数字相除:

\frac{2}{2} = 1

= 1 - 2

二次方程组的解是:

u = 1 + 2, u = 1 - 2

u = tan (x)

代回

tan (x) = 1 + 2, tan (x) = 1 - 2

tan (x) = 1 + 2, tan (x) = 1 - 2

tan (x) = 1 + 2 : x = arctan (1 + 2) + πn

tan (x) = 1 + 2

使用反三角函数性质

tan (x) = 1 + 2

tan (x) = 1 + 2

的通解

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

x = arctan (1 + 2) + πn

x = arctan (1 + 2) + πn

tan (x) = 1 - 2 : x = arctan (1 - 2) + πn

tan (x) = 1 - 2

使用反三角函数性质

tan (x) = 1 - 2

tan (x) = 1 - 2

的通解

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

x = arctan (1 - 2) + πn

x = arctan (1 - 2) + πn

合并所有解

x = arctan (1 + 2) + πn, x = arctan (1 - 2) + πn

以小数形式表示解

x = 1.17809 \dots + πn, x = - 0.39269 \dots + πn

作图

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