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受欢迎的三角函数 >

sec^2(x)+7tan(x)-9=0

解答

sec^{2} (x) + 7 tan (x) - 9 = 0

解答

x = \frac{π}{4} + πn, x = - 1.44644 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = - 82.87498 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) + 7 tan (x) - 9 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= - 9 + tan^{2} (x) + 1 + 7 tan (x)

化简

- 9 + tan^{2} (x) + 1 + 7 tan (x) : tan^{2} (x) + 7 tan (x) - 8

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

对同类项分组

= tan^{2} (x) + 7 tan (x) - 9 + 1

数字相加/相减:

- 9 + 1 = - 8

= tan^{2} (x) + 7 tan (x) - 8

= tan^{2} (x) + 7 tan (x) - 8

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

用替代法求解

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

令

: tan (x) = u

- 8 + u^{2} + 7 u = 0

- 8 + u^{2} + 7 u = 0 : u = 1, u = - 8

- 8 + u^{2} + 7 u = 0

改写成标准形式

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

u^{2} + 7 u - 8 = 0

使用求根公式求解

u^{2} + 7 u - 8 = 0

二次方程求根公式:

若

a = 1, b = 7, c = - 8

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 8) = 9

7^{2} - 4 \cdot 1 \cdot (- 8)

使用法则

- (- a) = a

= 7^{2} + 4 \cdot 1 \cdot 8

数字相乘:

4 \cdot 1 \cdot 8 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 7 + 9}{2 \cdot 1}, u_{2} = \frac{- 7 - 9}{2 \cdot 1}

u = \frac{- 7 + 9}{2 \cdot 1} : 1

\frac{- 7 + 9}{2 \cdot 1}

数字相加/相减:

- 7 + 9 = 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{- 7 - 9}{2 \cdot 1} : - 8

\frac{- 7 - 9}{2 \cdot 1}

数字相减:

- 7 - 9 = - 16

= \frac{- 16}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 16}{2}

使用分式法则:

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

= - \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= - 8

二次方程组的解是:

u = 1, u = - 8

u = tan (x)

代回

tan (x) = 1, tan (x) = - 8

tan (x) = 1, tan (x) = - 8

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) = - 8 : x = arctan (- 8) + πn

tan (x) = - 8

使用反三角函数性质

tan (x) = - 8

tan (x) = - 8

的通解

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

x = arctan (- 8) + πn

x = arctan (- 8) + πn

合并所有解

x = \frac{π}{4} + πn, x = arctan (- 8) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = - 1.44644 \dots + πn

作图

流行的例子

sin (x) = - 0.6

5+2sin^2(x)=3sin(x)+4

5 + 2 sin^{2} (x) = 3 sin (x) + 4

sin^2(2x)+cos(2x)=1

sin^{2} (2 x) + cos (2 x) = 1

sin (\frac{π}{x}) = 0

tan (x) = \frac{150}{80}