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受欢迎的三角函数 >

tan^2(θ)+sec(θ)=-2

解答

tan^{2} (θ) + sec (θ) = - 2

解答

θ \in R 无解

求解步骤

tan^{2} (θ) + sec (θ) = - 2

两边减去

- 2

tan^{2} (θ) + sec (θ) + 2 = 0

使用三角恒等式改写

2 + sec (θ) + tan^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= 2 + sec (θ) + sec^{2} (θ) - 1

化简

2 + sec (θ) + sec^{2} (θ) - 1 : sec^{2} (θ) + sec (θ) + 1

2 + sec (θ) + sec^{2} (θ) - 1

对同类项分组

= sec (θ) + sec^{2} (θ) + 2 - 1

数字相加/相减:

2 - 1 = 1

= sec^{2} (θ) + sec (θ) + 1

= sec^{2} (θ) + sec (θ) + 1

1 + sec (θ) + sec^{2} (θ) = 0

用替代法求解

1 + sec (θ) + sec^{2} (θ) = 0

令

: sec (θ) = u

1 + u + u^{2} = 0

1 + u + u^{2} = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{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}

化简

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

u_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 3 i}{2 \cdot 1}, u_{2} = \frac{- 1 - 3 i}{2 \cdot 1}

u = \frac{- 1 + 3 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 3 i}{2}

将

\frac{- 1 + 3 i}{2}

改写成标准复数形式:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

u = \frac{- 1 - 3 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 3 i}{2}

将

\frac{- 1 - 3 i}{2}

改写成标准复数形式:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3}{2} i

二次方程组的解是:

u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = sec (θ)

代回

sec (θ) = - \frac{1}{2} + i \frac{3}{2}, sec (θ) = - \frac{1}{2} - i \frac{3}{2}

sec (θ) = - \frac{1}{2} + i \frac{3}{2}, sec (θ) = - \frac{1}{2} - i \frac{3}{2}

sec (θ) = - \frac{1}{2} + i \frac{3}{2} :

无解

sec (θ) = - \frac{1}{2} + i \frac{3}{2}

无解

sec (θ) = - \frac{1}{2} - i \frac{3}{2} :

无解

sec (θ) = - \frac{1}{2} - i \frac{3}{2}

无解

合并所有解

θ \in R 无解

作图

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