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受欢迎的三角函数 >

6sec^2(x)+tan(x)-7=0

解答

6 sec^{2} (x) + tan (x) - 7 = 0

解答

x = 0.32175 \dots + πn, x = - 0.46364 \dots + πn

+1

度数

x = 18.43494 \dots^{\circ} + 18 0^{\circ} n, x = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

6 sec^{2} (x) + tan (x) - 7 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 7 + tan (x) + 6 (tan^{2} (x) + 1)

化简

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

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

乘开

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

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

使用分配律:

a (b + c) = ab + a c

a = 6, b = tan^{2} (x), c = 1

= 6 tan^{2} (x) + 6 \cdot 1

数字相乘:

6 \cdot 1 = 6

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

= - 7 + tan (x) + 6 tan^{2} (x) + 6

化简

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

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

对同类项分组

= tan (x) + 6 tan^{2} (x) - 7 + 6

数字相加/相减:

- 7 + 6 = - 1

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

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

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

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

用替代法求解

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

令

: tan (x) = u

- 1 + u + 6 u^{2} = 0

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

- 1 + u + 6 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 6, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 1) = 5

1^{2} - 4 \cdot 6 (- 1)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 6}, u_{2} = \frac{- 1 - 5}{2 \cdot 6}

u = \frac{- 1 + 5}{2 \cdot 6} : \frac{1}{3}

\frac{- 1 + 5}{2 \cdot 6}

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4}{12}

约分:

4

= \frac{1}{3}

u = \frac{- 1 - 5}{2 \cdot 6} : - \frac{1}{2}

\frac{- 1 - 5}{2 \cdot 6}

数字相减:

- 1 - 5 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

二次方程组的解是:

u = \frac{1}{3}, u = - \frac{1}{2}

u = tan (x)

代回

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{2}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{2}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

使用反三角函数性质

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3}

的通解

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{2} : x = arctan (- \frac{1}{2}) + πn

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

使用反三角函数性质

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

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

的通解

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

x = arctan (- \frac{1}{2}) + πn

x = arctan (- \frac{1}{2}) + πn

合并所有解

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{2}) + πn

以小数形式表示解

x = 0.32175 \dots + πn, x = - 0.46364 \dots + πn

作图

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