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受欢迎的三角函数 >

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

解答

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

解答

x = 1.10714 \dots + πn, x = 0.46364 \dots + πn

+1

度数

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 26.56505 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

5 tan (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

用替代法求解

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

令

: tan (x) = u

(1 + u^{2}) \cdot 2 - 5 u = 0

(1 + u^{2}) \cdot 2 - 5 u = 0 : u = 2, u = \frac{1}{2}

(1 + u^{2}) \cdot 2 - 5 u = 0

展开

(1 + u^{2}) \cdot 2 - 5 u : 2 + 2 u^{2} - 5 u

(1 + u^{2}) \cdot 2 - 5 u

= 2 (1 + u^{2}) - 5 u

乘开

2 (1 + u^{2}) : 2 + 2 u^{2}

2 (1 + u^{2})

使用分配律:

a (b + c) = ab + a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 + 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 + 2 u^{2}

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

2 + 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 + 2 = 0

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 5)^{2} - 4 \cdot 2 \cdot 2 = 3

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

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 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

u = \frac{- ( - 5 ) + 3}{2 \cdot 2} : 2

\frac{- ( - 5 ) + 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 + 3}{2 \cdot 2}

数字相加:

5 + 3 = 8

= \frac{8}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

u = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 5 ) - 3}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 - 3}{2 \cdot 2}

数字相减:

5 - 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = tan (x)

代回

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

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

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

使用反三角函数性质

tan (x) = 2

tan (x) = 2

的通解

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

x = arctan (2) + πn

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

以小数形式表示解

x = 1.10714 \dots + πn, x = 0.46364 \dots + πn

作图

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