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受欢迎的三角函数 >

sec^2(x)+tan^2(x)=5tan(x)

解答

sec^{2} (x) + tan^{2} (x) = 5 tan (x)

解答

x = 1.15759 \dots + πn, x = 0.21580 \dots + πn

+1

度数

x = 66.32508 \dots^{\circ} + 18 0^{\circ} n, x = 12.36498 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + tan^{2} (x) = 5 tan (x)

两边减去

5 tan (x)

sec^{2} (x) + tan^{2} (x) - 5 tan (x) = 0

使用三角恒等式改写

sec^{2} (x) + tan^{2} (x) - 5 tan (x)

使用毕达哥拉斯恒等式:

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

= tan^{2} (x) + 1 + tan^{2} (x) - 5 tan (x)

化简

tan^{2} (x) + 1 + tan^{2} (x) - 5 tan (x) : 2 tan^{2} (x) - 5 tan (x) + 1

tan^{2} (x) + 1 + tan^{2} (x) - 5 tan (x)

对同类项分组

= tan^{2} (x) + tan^{2} (x) - 5 tan (x) + 1

同类项相加:

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

= 2 tan^{2} (x) - 5 tan (x) + 1

= 2 tan^{2} (x) - 5 tan (x) + 1

1 + 2 tan^{2} (x) - 5 tan (x) = 0

用替代法求解

1 + 2 tan^{2} (x) - 5 tan (x) = 0

令

: tan (x) = u

1 + 2 u^{2} - 5 u = 0

1 + 2 u^{2} - 5 u = 0 : u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

1 + 2 u^{2} - 5 u = 0

改写成标准形式

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

2 u^{2} - 5 u + 1 = 0

使用求根公式求解

2 u^{2} - 5 u + 1 = 0

二次方程求根公式:

若

a = 2, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

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

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

数字相减:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

u = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

\frac{- ( - 5 ) + 17}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 + 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{5 + 17}{4}

u = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

\frac{- ( - 5 ) - 17}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 - 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{5 - 17}{4}

二次方程组的解是:

u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

u = tan (x)

代回

tan (x) = \frac{5 + 17}{4}, tan (x) = \frac{5 - 17}{4}

tan (x) = \frac{5 + 17}{4}, tan (x) = \frac{5 - 17}{4}

tan (x) = \frac{5 + 17}{4} : x = arctan (\frac{5 + 17}{4}) + πn

tan (x) = \frac{5 + 17}{4}

使用反三角函数性质

tan (x) = \frac{5 + 17}{4}

tan (x) = \frac{5 + 17}{4}

的通解

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

x = arctan (\frac{5 + 17}{4}) + πn

x = arctan (\frac{5 + 17}{4}) + πn

tan (x) = \frac{5 - 17}{4} : x = arctan (\frac{5 - 17}{4}) + πn

tan (x) = \frac{5 - 17}{4}

使用反三角函数性质

tan (x) = \frac{5 - 17}{4}

tan (x) = \frac{5 - 17}{4}

的通解

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

x = arctan (\frac{5 - 17}{4}) + πn

x = arctan (\frac{5 - 17}{4}) + πn

合并所有解

x = arctan (\frac{5 + 17}{4}) + πn, x = arctan (\frac{5 - 17}{4}) + πn

以小数形式表示解

x = 1.15759 \dots + πn, x = 0.21580 \dots + πn

作图

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