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受欢迎的三角函数 >

tan(x)=sec^2(x)

解答

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

解答

x \in R 无解

求解步骤

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

两边减去

sec^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

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

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

用替代法求解

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

令

: tan (x) = 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 ( - 1 ) ( - 1 )}{2 ( - 1 )}

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

化简

1^{2} - 4 (- 1) (- 1) : 3 i

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 1)

使用法则

- (- a) = a

= 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 ( - 1 )}

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

= - \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})

去除括号:

(a) = a, - (- a) = a

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

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

= - \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})

使用法则

- (- a) = a

= \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 = tan (x)

代回

tan (x) = \frac{1}{2} - i \frac{3}{2}, tan (x) = \frac{1}{2} + i \frac{3}{2}

tan (x) = \frac{1}{2} - i \frac{3}{2}, tan (x) = \frac{1}{2} + i \frac{3}{2}

tan (x) = \frac{1}{2} - i \frac{3}{2} :

无解

tan (x) = \frac{1}{2} - i \frac{3}{2}

无解

tan (x) = \frac{1}{2} + i \frac{3}{2} :

无解

tan (x) = \frac{1}{2} + i \frac{3}{2}

无解

合并所有解

x \in R 无解

作图

流行的例子

sin (x) = 0, 8

sin(x)tan(x)=-2-cot(x)sin(x)

sin (x) tan (x) = - 2 - cot (x) sin (x)

-2sin^2(x)+cos(x)(2+2cos(x))=0

- 2 sin^{2} (x) + cos (x) (2 + 2 cos (x)) = 0

tan (\frac{θ}{2}) = 2

sin (x) = 6