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受欢迎的三角函数 >

3sec^2(x)+tan(x)-6=0

解答

3 sec^{2} (x) + tan (x) - 6 = 0

解答

x = 0.70282 \dots + πn, x = - 0.86797 \dots + πn

+1

度数

x = 40.26883 \dots^{\circ} + 18 0^{\circ} n, x = - 49.73116 \dots^{\circ} + 18 0^{\circ} n

求解步骤

3 sec^{2} (x) + tan (x) - 6 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 6 + tan (x) + 3 (tan^{2} (x) + 1)

化简

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

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

乘开

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

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

使用分配律:

a (b + c) = ab + a c

a = 3, b = tan^{2} (x), c = 1

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

数字相乘:

3 \cdot 1 = 3

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

= - 6 + tan (x) + 3 tan^{2} (x) + 3

化简

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

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

对同类项分组

= tan (x) + 3 tan^{2} (x) - 6 + 3

数字相加/相减:

- 6 + 3 = - 3

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

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

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

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

用替代法求解

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

令

: tan (x) = u

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

- 3 + u + 3 u^{2} = 0 : u = \frac{- 1 + 37}{6}, u = \frac{- 1 - 37}{6}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 \cdot 3 (- 3) = 37

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 3)

使用法则

- (- a) = a

= 1 + 4 \cdot 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 1 + 36

数字相加:

1 + 36 = 37

= 37

u_{1, 2} = \frac{- 1 \pm 37}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 1 + 37}{2 \cdot 3}, u_{2} = \frac{- 1 - 37}{2 \cdot 3}

u = \frac{- 1 + 37}{2 \cdot 3} : \frac{- 1 + 37}{6}

\frac{- 1 + 37}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 1 + 37}{6}

u = \frac{- 1 - 37}{2 \cdot 3} : \frac{- 1 - 37}{6}

\frac{- 1 - 37}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 1 - 37}{6}

二次方程组的解是:

u = \frac{- 1 + 37}{6}, u = \frac{- 1 - 37}{6}

u = tan (x)

代回

tan (x) = \frac{- 1 + 37}{6}, tan (x) = \frac{- 1 - 37}{6}

tan (x) = \frac{- 1 + 37}{6}, tan (x) = \frac{- 1 - 37}{6}

tan (x) = \frac{- 1 + 37}{6} : x = arctan (\frac{- 1 + 37}{6}) + πn

tan (x) = \frac{- 1 + 37}{6}

使用反三角函数性质

tan (x) = \frac{- 1 + 37}{6}

tan (x) = \frac{- 1 + 37}{6}

的通解

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

x = arctan (\frac{- 1 + 37}{6}) + πn

x = arctan (\frac{- 1 + 37}{6}) + πn

tan (x) = \frac{- 1 - 37}{6} : x = arctan (\frac{- 1 - 37}{6}) + πn

tan (x) = \frac{- 1 - 37}{6}

使用反三角函数性质

tan (x) = \frac{- 1 - 37}{6}

tan (x) = \frac{- 1 - 37}{6}

的通解

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

x = arctan (\frac{- 1 - 37}{6}) + πn

x = arctan (\frac{- 1 - 37}{6}) + πn

合并所有解

x = arctan (\frac{- 1 + 37}{6}) + πn, x = arctan (\frac{- 1 - 37}{6}) + πn

以小数形式表示解

x = 0.70282 \dots + πn, x = - 0.86797 \dots + πn

作图

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