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受欢迎的三角函数 >

sec^2(x)-6tan(x)=-4

解答

sec^{2} (x) - 6 tan (x) = - 4

解答

x = 1.37340 \dots + πn, x = \frac{π}{4} + πn

+1

度数

x = 78.69006 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (x) - 6 tan (x) = - 4

两边减去

- 4

sec^{2} (x) - 6 tan (x) + 4 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= 4 + tan^{2} (x) + 1 - 6 tan (x)

化简

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

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

对同类项分组

= tan^{2} (x) - 6 tan (x) + 4 + 1

数字相加:

4 + 1 = 5

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

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

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

用替代法求解

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

令

: tan (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

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

数字相乘:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

数字相减:

36 - 20 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

4^{2} = 4

= 4

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

将解分隔开

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

u = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

\frac{- ( - 6 ) + 4}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{6 + 4}{2 \cdot 1}

数字相加:

6 + 4 = 10

= \frac{10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

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

使用法则

- (- a) = a

= \frac{6 - 4}{2 \cdot 1}

数字相减:

6 - 4 = 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

二次方程组的解是:

u = 5, u = 1

u = tan (x)

代回

tan (x) = 5, tan (x) = 1

tan (x) = 5, tan (x) = 1

tan (x) = 5 : x = arctan (5) + πn

tan (x) = 5

使用反三角函数性质

tan (x) = 5

tan (x) = 5

的通解

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

x = arctan (5) + πn

x = arctan (5) + πn

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

合并所有解

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

以小数形式表示解

x = 1.37340 \dots + πn, x = \frac{π}{4} + πn

作图

流行的例子

64cos^2(x)-16=0

cos(2x)=-(sqrt(2))/2

tan(θ/2)=sqrt(3)