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 >

2cos(x/2)-sqrt(2)=0,0<= x<2pi

解答

2 cos (\frac{x}{2}) - 2 = 0, 0 \leq x < 2 π

解答

x = \frac{π}{2}

+1

度数

x = 9 0^{\circ}

求解步骤

2 cos (\frac{x}{2}) - 2 = 0, 0 \leq x < 2 π

将

2

到右边

2 cos (\frac{x}{2}) - 2 = 0

两边加上

2

2 cos (\frac{x}{2}) - 2 + 2 = 0 + 2

化简

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

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

两边除以

2

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

两边除以

2

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

化简

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

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

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

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}

\frac{x}{2} = \frac{π}{4} + 2 πn, \frac{x}{2} = \frac{7 π}{4} + 2 πn

\frac{x}{2} = \frac{π}{4} + 2 πn, \frac{x}{2} = \frac{7 π}{4} + 2 πn

解

\frac{x}{2} = \frac{π}{4} + 2 πn : x = \frac{π}{2} + 4 πn

\frac{x}{2} = \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{4} + 2 \cdot 2 πn : \frac{π}{2} + 4 πn

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

解

\frac{x}{2} = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{2} + 4 πn

\frac{x}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn : \frac{7 π}{2} + 4 πn

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{4} = \frac{7 π}{2}

2 \cdot \frac{7 π}{4}

分式相乘:

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

= \frac{7 π 2}{4}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{4}

约分:

2

= \frac{7 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

x = \frac{7 π}{2} + 4 πn

x = \frac{π}{2} + 4 πn, x = \frac{7 π}{2} + 4 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}

作图

流行的例子

cos(x/4)=-(sqrt(2))/2

cos (\frac{x}{4}) = - \frac{2}{2}

tan (x) = \frac{12}{20}

15 cos (x) = 0

sin(pi/2-x)=cos^2(x)

sin (\frac{π}{2} - x) = cos^{2} (x)

tan(x)sin(x)+sin(-x)=0,0<= x<= 2pi

tan (x) sin (x) + sin (- x) = 0, 0 \leq x \leq 2 π