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 >

3arccos(5x)=2pi

解答

3 arccos (5 x) = 2 π

解答

x = - \frac{1}{10}

求解步骤

3 arccos (5 x) = 2 π

两边除以

3

3 arccos (5 x) = 2 π

两边除以

3

\frac{3 arccos ( 5 x )}{3} = \frac{2 π}{3}

化简

arccos (5 x) = \frac{2 π}{3}

arccos (5 x) = \frac{2 π}{3}

使用反三角函数性质

arccos (5 x) = \frac{2 π}{3}

arccos (x) = a \Rightarrow x = cos (a)

5 x = cos (\frac{2 π}{3})

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

使用以下普通恒等式:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

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{1}{2}

= - \frac{1}{2}

5 x = - \frac{1}{2}

5 x = - \frac{1}{2}

解

5 x = - \frac{1}{2} : x = - \frac{1}{10}

5 x = - \frac{1}{2}

两边除以

5

5 x = - \frac{1}{2}

两边除以

5

\frac{5 x}{5} = \frac{- \frac{1}{2}}{5}

化简

\frac{5 x}{5} = \frac{- \frac{1}{2}}{5}

化简

\frac{5 x}{5} : x

\frac{5 x}{5}

数字相除:

\frac{5}{5} = 1

= x

化简

\frac{- \frac{1}{2}}{5} : - \frac{1}{10}

\frac{- \frac{1}{2}}{5}

使用分式法则:

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

= - \frac{\frac{1}{2}}{5}

使用分式法则:

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

\frac{\frac{1}{2}}{5} = \frac{1}{2 \cdot 5}

= - \frac{1}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= - \frac{1}{10}

x = - \frac{1}{10}

x = - \frac{1}{10}

x = - \frac{1}{10}

x = - \frac{1}{10}

作图

流行的例子

sqrt(3)cot(3θ+pi/3)+1=0

3 cot (3 θ + \frac{π}{3}) + 1 = 0

tan (θ) = \frac{4}{13}

2cos^2(x)-cos(x)-1=0,0<= x<= 2pi

2 cos^{2} (x) - cos (x) - 1 = 0, 0 \leq x \leq 2 π

sin (θ) = - \frac{6}{7}

asin(θ)=bcos(θ)

a sin (θ) = b cos (θ)