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受欢迎的三角函数 >

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

解答

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

解答

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

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

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

在两边乘以

1 - u

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

在两边乘以

1 - u

\frac{u ^{2}}{1 - u} (1 - u) = 1 \cdot (1 - u) + u (1 - u)

化简

\frac{u ^{2}}{1 - u} (1 - u) = 1 \cdot (1 - u) + u (1 - u)

化简

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

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

分式相乘:

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

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

约分:

1 - u

= u^{2}

化简

1 \cdot (1 - u) : 1 - u

1 \cdot (1 - u)

乘以:

1 \cdot (1 - u) = (1 - u)

= (1 - u)

去除括号:

(a) = a

= 1 - u

u^{2} = 1 - u + u (1 - u)

u^{2} = 1 - u + u (1 - u)

u^{2} = 1 - u + u (1 - u)

解

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

u^{2} = 1 - u + u (1 - u)

展开

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

1 - u + u (1 - u)

乘开

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

u (1 - u)

使用分配律:

a (b - c) = ab - a c

a = u, b = 1, c = u

= u \cdot 1 - uu

= 1 \cdot u - uu

化简

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

1 \cdot u - uu

1 \cdot u = u

1 \cdot u

乘以:

1 \cdot u = u

= u

uu = u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

= u - u^{2}

= u - u^{2}

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

同类项相加:

- u + u = 0

= 1 - u^{2}

u^{2} = 1 - u^{2}

将

u^{2}

para o lado esquerdo

u^{2} = 1 - u^{2}

两边加上

u^{2}

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

化简

2 u^{2} = 1

2 u^{2} = 1

两边除以

2

2 u^{2} = 1

两边除以

2

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

化简

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

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

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

验证解

找到无定义的点（奇点）:

u = 1

取

\frac{u ^{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

两边除以

- 1

- u = - 1

两边除以

- 1

\frac{- u}{- 1} = \frac{- 1}{- 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 = \frac{1}{2}, u = - \frac{1}{2}

u = cos (x)

代回

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

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

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

使用反三角函数性质

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

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

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

合并所有解

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

以小数形式表示解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

作图

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