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 >

sin(x)=(-(6))/((7)cos(x))

解答

sin (x) = \frac{- ( 6 )}{( 7 ) cos ( x )}

解答

x \in R 无解

求解步骤

sin (x) = \frac{- ( 6 )}{( 7 ) cos ( x )}

两边减去

\frac{- 6}{7 cos ( x )}

sin (x) + \frac{6}{7 cos ( x )} = 0

化简

sin (x) + \frac{6}{7 cos ( x )} : \frac{7 sin ( x ) cos ( x ) + 6}{7 cos ( x )}

sin (x) + \frac{6}{7 cos ( x )}

将项转换为分式:

sin (x) = \frac{sin ( x ) 7 cos ( x )}{7 cos ( x )}

= \frac{sin ( x ) \cdot 7 cos ( x )}{7 cos ( x )} + \frac{6}{7 cos ( x )}

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{sin ( x ) \cdot 7 cos ( x ) + 6}{7 cos ( x )}

\frac{7 sin ( x ) cos ( x ) + 6}{7 cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

7 sin (x) cos (x) + 6 = 0

使用三角恒等式改写

7 sin (x) cos (x) + 6

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= 6 + 7 \cdot \frac{sin ( 2 x )}{2}

6 + 7 \cdot \frac{sin ( 2 x )}{2} = 0

将

6

到右边

6 + 7 \cdot \frac{sin ( 2 x )}{2} = 0

两边减去

6

6 + 7 \cdot \frac{sin ( 2 x )}{2} - 6 = 0 - 6

化简

7 \cdot \frac{sin ( 2 x )}{2} = - 6

7 \cdot \frac{sin ( 2 x )}{2} = - 6

整理

7 \cdot \frac{sin ( 2 x )}{2} : \frac{7 sin ( 2 x )}{2}

7 \cdot \frac{sin ( 2 x )}{2}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 7}{2}

\frac{7 sin ( 2 x )}{2} = - 6

在两边乘以

2

\frac{7 sin ( 2 x )}{2} = - 6

在两边乘以

2

\frac{7 sin ( 2 x )}{2} \cdot 2 = - 6 \cdot 2

化简

\frac{7 sin ( 2 x )}{2} \cdot 2 = - 6 \cdot 2

化简

\frac{7 sin ( 2 x )}{2} \cdot 2 : 7 sin (2 x)

\frac{7 sin ( 2 x )}{2} \cdot 2

分式相乘:

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

= \frac{7 sin ( 2 x ) \cdot 2}{2}

约分:

2

= 7 sin (2 x)

化简

- 6 \cdot 2 : - 12

- 6 \cdot 2

数字相乘:

6 \cdot 2 = 12

= - 12

7 sin (2 x) = - 12

7 sin (2 x) = - 12

7 sin (2 x) = - 12

两边除以

7

7 sin (2 x) = - 12

两边除以

7

\frac{7 sin ( 2 x )}{7} = \frac{- 12}{7}

化简

sin (2 x) = - \frac{12}{7}

sin (2 x) = - \frac{12}{7}

- 1 \leq sin (x) \leq 1

x \in R 无解

作图

流行的例子

5^2=8^2-(11.85)^2-2(8)(11.85)(cos(A))

5^{2} = 8^{2} - (11.85)^{2} - 2 (8) (11.85) (cos (A))

2 + 2 sin (θ) = 4

- 4 sin (8 x) = 0

-sin(x)+2cos(x)=0

- sin (x) + 2 cos (x) = 0

(sin(θ))/(140)=(sin(110))/(622)

\frac{sin ( θ )}{140} = \frac{sin ( 11 0 ^{\circ} )}{622}