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受欢迎的三角函数 >

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

解答

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

解答

x = πn, x = \frac{π}{4} + πn

+1

度数

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

求解步骤

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

两边减去

sec^{2} (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: tan (x) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

改写成标准形式

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

- u^{2} + u = 0

使用求根公式求解

- u^{2} + u = 0

二次方程求根公式:

若

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = tan (x)

代回

tan (x) = 0, tan (x) = 1

tan (x) = 0, tan (x) = 1

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

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 = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = 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 = πn, x = \frac{π}{4} + πn

作图

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