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受欢迎的三角函数 >

cos(x)-sqrt(cos(x))=0

解答

cos (x) - cos (x) = 0

解答

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

+1

度数

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

求解步骤

cos (x) - cos (x) = 0

用替代法求解

cos (x) - cos (x) = 0

令

: cos (x) = u

u - u = 0

u - u = 0 : u = 1, u = 0

u - u = 0

去除平方根

u - u = 0

两边减去

u

u - u - u = 0 - u

化简

- u = - u

两边进行平方:

u = u^{2}

u - u = 0

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

展开

(- u)^{2} : u

(- u)^{2}

使用指数法则:

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

若

n

是偶数

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

= (u)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= u^{\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

= u

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}

u = u^{2}

u = u^{2}

u = u^{2}

解

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

u = u^{2}

交换两边

u^{2} = u

将

u

para o lado esquerdo

u^{2} = u

两边减去

u

u^{2} - u = u - u

化简

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

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

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

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

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

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

二次方程组的解是:

u = 1, u = 0

u = 1, u = 0

验证解:

u = 1

真

, u = 0

真

将它们代入

u - u = 0

检验解是否符合

去除与方程不符的解。

代入

u = 1 :

真

1 - 1 = 0

1 - 1 = 0

1 - 1

使用法则

1 = 1

= 1 - 1

数字相减:

1 - 1 = 0

= 0

0 = 0

真

代入

u = 0 :

真

0 - 0 = 0

0 - 0 = 0

0 - 0

使用法则

0 = 0

= 0 - 0

数字相减:

0 - 0 = 0

= 0

0 = 0

真

解为

u = 1, u = 0

u = cos (x)

代回

cos (x) = 1, cos (x) = 0

cos (x) = 1, cos (x) = 0

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) = 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

合并所有解

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

作图

流行的例子

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