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受欢迎的三角函数 >

sec^2(x)+5tan(x)=7

解答

sec^{2} (x) + 5 tan (x) = 7

解答

x = \frac{π}{4} + πn, x = - 1.40564 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 80.53767 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + 5 tan (x) = 7

两边减去

7

sec^{2} (x) + 5 tan (x) - 7 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 7 + tan^{2} (x) + 1 + 5 tan (x)

化简

- 7 + tan^{2} (x) + 1 + 5 tan (x) : tan^{2} (x) + 5 tan (x) - 6

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

对同类项分组

= tan^{2} (x) + 5 tan (x) - 7 + 1

数字相加/相减:

- 7 + 1 = - 6

= tan^{2} (x) + 5 tan (x) - 6

= tan^{2} (x) + 5 tan (x) - 6

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

用替代法求解

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

令

: tan (x) = u

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

- 6 + u^{2} + 5 u = 0 : u = 1, u = - 6

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

改写成标准形式

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

u^{2} + 5 u - 6 = 0

使用求根公式求解

u^{2} + 5 u - 6 = 0

二次方程求根公式:

若

a = 1, b = 5, c = - 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot (- 6) = 7

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

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 \cdot 1}, u_{2} = \frac{- 5 - 7}{2 \cdot 1}

u = \frac{- 5 + 7}{2 \cdot 1} : 1

\frac{- 5 + 7}{2 \cdot 1}

数字相加/相减:

- 5 + 7 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- 5 - 7}{2 \cdot 1} : - 6

\frac{- 5 - 7}{2 \cdot 1}

数字相减:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 12}{2}

使用分式法则:

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

= - \frac{12}{2}

数字相除:

\frac{12}{2} = 6

= - 6

二次方程组的解是:

u = 1, u = - 6

u = tan (x)

代回

tan (x) = 1, tan (x) = - 6

tan (x) = 1, tan (x) = - 6

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 6 : x = arctan (- 6) + πn

tan (x) = - 6

使用反三角函数性质

tan (x) = - 6

tan (x) = - 6

的通解

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

x = arctan (- 6) + πn

x = arctan (- 6) + πn

合并所有解

x = \frac{π}{4} + πn, x = arctan (- 6) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.40564 \dots + πn

作图

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sin(2x)=-5cos(2x)

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