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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

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

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

在两边乘以

1 + u

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

在两边乘以

1 + u

2 (1 + u) + 2 u (1 + u) = \frac{2}{1 + u} (1 + u)

化简

2 (1 + u) + 2 u (1 + u) = 2

2 (1 + u) + 2 u (1 + u) = 2

解

2 (1 + u) + 2 u (1 + u) = 2 : u = 0, u = - 2

2 (1 + u) + 2 u (1 + u) = 2

展开

2 (1 + u) + 2 u (1 + u) : 2 + 4 u + 2 u^{2}

2 (1 + u) + 2 u (1 + u)

乘开

2 (1 + u) : 2 + 2 u

2 (1 + u)

使用分配律:

a (b + c) = ab + a c

a = 2, b = 1, c = u

= 2 \cdot 1 + 2 u

数字相乘:

2 \cdot 1 = 2

= 2 + 2 u

= 2 + 2 u + 2 u (1 + u)

乘开

2 u (1 + u) : 2 u + 2 u^{2}

2 u (1 + u)

使用分配律:

a (b + c) = ab + a c

a = 2 u, b = 1, c = u

= 2 u \cdot 1 + 2 uu

= 2 \cdot 1 \cdot u + 2 uu

化简

2 \cdot 1 \cdot u + 2 uu : 2 u + 2 u^{2}

2 \cdot 1 \cdot u + 2 uu

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

数字相乘:

2 \cdot 1 = 2

= 2 u

2 uu = 2 u^{2}

2 uu

使用指数法则:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= 2 u^{2}

= 2 u + 2 u^{2}

= 2 u + 2 u^{2}

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

同类项相加:

2 u + 2 u = 4 u

= 2 + 4 u + 2 u^{2}

2 + 4 u + 2 u^{2} = 2

将

2

para o lado esquerdo

2 + 4 u + 2 u^{2} = 2

两边减去

2

2 + 4 u + 2 u^{2} - 2 = 2 - 2

化简

2 u^{2} + 4 u = 0

2 u^{2} + 4 u = 0

使用求根公式求解

2 u^{2} + 4 u = 0

二次方程求根公式:

若

a = 2, b = 4, c = 0

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

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

4^{2} - 4 \cdot 2 \cdot 0 = 4

4^{2} - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

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

将解分隔开

u_{1} = \frac{- 4 + 4}{2 \cdot 2}, u_{2} = \frac{- 4 - 4}{2 \cdot 2}

u = \frac{- 4 + 4}{2 \cdot 2} : 0

\frac{- 4 + 4}{2 \cdot 2}

数字相加/相减:

- 4 + 4 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

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

= 0

u = \frac{- 4 - 4}{2 \cdot 2} : - 2

\frac{- 4 - 4}{2 \cdot 2}

数字相减:

- 4 - 4 = - 8

= \frac{- 8}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

二次方程组的解是:

u = 0, u = - 2

u = 0, u = - 2

验证解

找到无定义的点（奇点）:

u = - 1

取

\frac{2}{1 + u}

的分母，令其等于零

解

1 + u = 0 : u = - 1

1 + u = 0

将

1

到右边

1 + u = 0

两边减去

1

1 + u - 1 = 0 - 1

化简

u = - 1

u = - 1

以下点无定义

u = - 1

将不在定义域的点与解相综合:

u = 0, u = - 2

u = cos (x)

代回

cos (x) = 0, cos (x) = - 2

cos (x) = 0, cos (x) = - 2

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - 2 :

无解

cos (x) = - 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

流行的例子

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