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受欢迎的三角函数 >

sec^2(b)=2+tan(b)

解答

sec^{2} (b) = 2 + tan (b)

解答

b = 1.01722 \dots + πn, b = - 0.55357 \dots + πn

+1

度数

b = 58.28252 \dots^{\circ} + 18 0^{\circ} n, b = - 31.71747 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (b) = 2 + tan (b)

两边减去

2 + tan (b)

sec^{2} (b) - 2 - tan (b) = 0

使用三角恒等式改写

- 2 + sec^{2} (b) - tan (b)

使用毕达哥拉斯恒等式:

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

= - 2 + tan^{2} (b) + 1 - tan (b)

化简

- 2 + tan^{2} (b) + 1 - tan (b) : tan^{2} (b) - tan (b) - 1

- 2 + tan^{2} (b) + 1 - tan (b)

对同类项分组

= tan^{2} (b) - tan (b) - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

= tan^{2} (b) - tan (b) - 1

= tan^{2} (b) - tan (b) - 1

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

用替代法求解

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

令

: tan (b) = u

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

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

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

改写成标准形式

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

u^{2} - u - 1 = 0

使用求根公式求解

u^{2} - u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 1

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

\frac{- ( - 1 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = tan (b)

代回

tan (b) = \frac{1 + 5}{2}, tan (b) = \frac{1 - 5}{2}

tan (b) = \frac{1 + 5}{2}, tan (b) = \frac{1 - 5}{2}

tan (b) = \frac{1 + 5}{2} : b = arctan (\frac{1 + 5}{2}) + πn

tan (b) = \frac{1 + 5}{2}

使用反三角函数性质

tan (b) = \frac{1 + 5}{2}

tan (b) = \frac{1 + 5}{2}

的通解

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

b = arctan (\frac{1 + 5}{2}) + πn

b = arctan (\frac{1 + 5}{2}) + πn

tan (b) = \frac{1 - 5}{2} : b = arctan (\frac{1 - 5}{2}) + πn

tan (b) = \frac{1 - 5}{2}

使用反三角函数性质

tan (b) = \frac{1 - 5}{2}

tan (b) = \frac{1 - 5}{2}

的通解

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

b = arctan (\frac{1 - 5}{2}) + πn

b = arctan (\frac{1 - 5}{2}) + πn

合并所有解

b = arctan (\frac{1 + 5}{2}) + πn, b = arctan (\frac{1 - 5}{2}) + πn

以小数形式表示解

b = 1.01722 \dots + πn, b = - 0.55357 \dots + πn

作图

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