zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

cos(x)= 1/(cos^2(x))

解答

cos (x) = \frac{1}{cos ^{2} ( x )}

解答

x = 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (x) = \frac{1}{cos ^{2} ( x )}

用替代法求解

cos (x) = \frac{1}{cos ^{2} ( x )}

令

: cos (x) = u

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

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

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

u u^{2} : u^{3}

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

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= u^{1 + 2}

数字相加:

1 + 2 = 3

= u^{3}

u^{3} = 1

u^{3} = 1

解

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

u^{3} = 1

对于

x^{3} = f (a)

解为

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

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

化简

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

化简

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1}{u ^{2}}

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = cos (x)

代回

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

的通解

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

无解

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

无解

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

无解

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

无解

合并所有解

x = 2 πn

作图

流行的例子

2cos(t)-1=0,0<,=t<= 2pi

2 cos (t) - 1 = 0, 0 <, = t \leq 2 π

2sin(2(θ+pi/6))=-sqrt(2)

2 sin (2 (θ + \frac{π}{6})) = - 2

tan (x) = 0.2092

sin(θ+1)=cos(θ)

sin (θ + 1) = cos (θ)

0 = sin (x + c)