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受欢迎的三角函数 >

tan(x)-sqrt(1-2tan^2(x))=0

解答

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

解答

x = 0.52359 \dots + πn

+1

度数

x = 3 0^{\circ} + 18 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: tan (x) = u

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

u - 1 - 2 u^{2} = 0 : u = \frac{1}{3}

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

去除平方根

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

两边减去

u

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

化简

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

两边进行平方:

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

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

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

展开

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

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

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

使用根式运算法则:

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

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

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 1 - 2 u^{2}

展开

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

(- u)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= u^{2}

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

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

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

解

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

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

将

1

到右边

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

两边减去

1

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

化简

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

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

将

u^{2}

para o lado esquerdo

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

两边减去

u^{2}

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

化简

- 3 u^{2} = - 1

- 3 u^{2} = - 1

两边除以

- 3

- 3 u^{2} = - 1

两边除以

- 3

\frac{- 3 u ^{2}}{- 3} = \frac{- 1}{- 3}

化简

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{3}, u = - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

验证解:

u = \frac{1}{3}

真

, u = - \frac{1}{3}

假

将它们代入

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

检验解是否符合

去除与方程不符的解。

代入

u = \frac{1}{3} :

真

\frac{1}{3} - 1 - 2 (\frac{1}{3})^{2} = 0

\frac{1}{3} - 1 - 2 (\frac{1}{3})^{2} = 0

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

1 - 2 (\frac{1}{3})^{2} = \frac{1}{3}

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

2 (\frac{1}{3})^{2} = \frac{2}{3}

2 (\frac{1}{3})^{2}

(\frac{1}{3})^{2} = \frac{1}{3}

(\frac{1}{3})^{2}

使用根式运算法则:

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

= ((\frac{1}{3})^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{1}{3})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= \frac{1}{3}

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{3}

= 1 - \frac{2}{3}

化简

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

1 - \frac{2}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

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

因为分母相等，所以合并分式:

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

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

1 \cdot 3 - 2 = 1

1 \cdot 3 - 2

数字相乘:

1 \cdot 3 = 3

= 3 - 2

数字相减:

3 - 2 = 1

= 1

= \frac{1}{3}

= \frac{1}{3}

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

同类项相加:

\frac{1}{3} - \frac{1}{3} = 0

= 0

0 = 0

真

代入

u = - \frac{1}{3} :

假

(- \frac{1}{3}) - 1 - 2 (- \frac{1}{3})^{2} = 0

(- \frac{1}{3}) - 1 - 2 (- \frac{1}{3})^{2} = - 2 \frac{1}{3}

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

去除括号:

(- a) = - a

= - \frac{1}{3} - 1 - 2 (- \frac{1}{3})^{2}

1 - 2 (- \frac{1}{3})^{2} = \frac{1}{3}

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

2 (- \frac{1}{3})^{2} = \frac{2}{3}

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

(- \frac{1}{3})^{2} = \frac{1}{3}

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- \frac{1}{3})^{2} = (\frac{1}{3})^{2}

= (\frac{1}{3})^{2}

使用根式运算法则:

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

= ((\frac{1}{3})^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= (\frac{1}{3})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= \frac{1}{3}

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{3}

= 1 - \frac{2}{3}

化简

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

1 - \frac{2}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

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

因为分母相等，所以合并分式:

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

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

1 \cdot 3 - 2 = 1

1 \cdot 3 - 2

数字相乘:

1 \cdot 3 = 3

= 3 - 2

数字相减:

3 - 2 = 1

= 1

= \frac{1}{3}

= \frac{1}{3}

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

同类项相加:

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

= - 2 \frac{1}{3}

- 2 \frac{1}{3} = 0

假

解是

u = \frac{1}{3}

u = tan (x)

代回

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

使用反三角函数性质

tan (x) = \frac{1}{3}

tan (x) = \frac{1}{3}

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

合并所有解

x = arctan (\frac{1}{3}) + πn

以小数形式表示解

x = 0.52359 \dots + πn

作图

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