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 >

-cos(x)= 2/(sqrt(20))

解答

- cos (x) = \frac{2}{20}

解答

x = 2.03444 \dots + 2 πn, x = - 2.03444 \dots + 2 πn

+1

度数

x = 116.56505 \dots^{\circ} + 36 0^{\circ} n, x = - 116.56505 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- cos (x) = \frac{2}{20}

20 = 25

20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

- cos (x) = \frac{2}{2 5}

数字相除:

\frac{2}{2} = 1

- cos (x) = \frac{1}{5}

两边除以

- 1

- cos (x) = \frac{1}{5}

两边除以

- 1

\frac{- cos ( x )}{- 1} = \frac{\frac{1}{5}}{- 1}

化简

\frac{- cos ( x )}{- 1} = \frac{\frac{1}{5}}{- 1}

化简

\frac{- cos ( x )}{- 1} : cos (x)

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

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= cos (x)

化简

\frac{\frac{1}{5}}{- 1} : - \frac{5}{5}

\frac{\frac{1}{5}}{- 1}

使用分式法则:

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

= - \frac{\frac{1}{5}}{1}

使用分式法则:

\frac{a}{1} = a

\frac{\frac{1}{5}}{1} = \frac{1}{5}

= - \frac{1}{5}

- \frac{1}{5}

有理化:

- \frac{5}{5}

- \frac{1}{5}

乘以共轭根式

\frac{5}{5}

= - \frac{1 \cdot 5}{5 5}

1 \cdot 5 = 5

55 = 5

55

使用根式运算法则:

a a = a

55 = 5

= 5

= - \frac{5}{5}

= - \frac{5}{5}

cos (x) = - \frac{5}{5}

cos (x) = - \frac{5}{5}

cos (x) = - \frac{5}{5}

使用反三角函数性质

cos (x) = - \frac{5}{5}

cos (x) = - \frac{5}{5}

的通解

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

x = arccos (- \frac{5}{5}) + 2 πn, x = - arccos (- \frac{5}{5}) + 2 πn

x = arccos (- \frac{5}{5}) + 2 πn, x = - arccos (- \frac{5}{5}) + 2 πn

以小数形式表示解

x = 2.03444 \dots + 2 πn, x = - 2.03444 \dots + 2 πn

作图

流行的例子

tan^2(x)+7=-10sec(x),0<x<2pi

tan^{2} (x) + 7 = - 10 sec (x), 0 < x < 2 π

tan (θ) = \frac{14}{11}

tan(3φ-4)=-1/2 ,0<= φ<= pi/2

tan (3 φ - 4) = - \frac{1}{2}, 0 \leq φ \leq \frac{π}{2}

solvefor g,θ=arctan((5.54)/(10g)+0.2)

so l v e f or g, θ = arctan (\frac{5.54}{10 g} + 0.2)

tan (t) = \frac{3}{4}