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 >

4sin(2x)tan(x)-tan(x)=0

解答

4 sin (2 x) tan (x) - tan (x) = 0

解答

x = πn, x = \frac{0.25268 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 7.23875 \dots^{\circ} + 18 0^{\circ} n, x = 82.76124 \dots^{\circ} + 18 0^{\circ} n

求解步骤

4 sin (2 x) tan (x) - tan (x) = 0

分解

4 sin (2 x) tan (x) - tan (x) : tan (x) (4 sin (2 x) - 1)

4 sin (2 x) tan (x) - tan (x)

因式分解出通项

tan (x)

= tan (x) (4 sin (2 x) - 1)

tan (x) (4 sin (2 x) - 1) = 0

分别求解每个部分

tan (x) = 0 or 4 sin (2 x) - 1 = 0

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

4 sin (2 x) - 1 = 0 : x = \frac{arcsin ( \frac{1}{4} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

4 sin (2 x) - 1 = 0

将

1

到右边

4 sin (2 x) - 1 = 0

两边加上

1

4 sin (2 x) - 1 + 1 = 0 + 1

化简

4 sin (2 x) = 1

4 sin (2 x) = 1

两边除以

4

4 sin (2 x) = 1

两边除以

4

\frac{4 sin ( 2 x )}{4} = \frac{1}{4}

化简

sin (2 x) = \frac{1}{4}

sin (2 x) = \frac{1}{4}

使用反三角函数性质

sin (2 x) = \frac{1}{4}

sin (2 x) = \frac{1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 x = arcsin (\frac{1}{4}) + 2 πn, 2 x = π - arcsin (\frac{1}{4}) + 2 πn

2 x = arcsin (\frac{1}{4}) + 2 πn, 2 x = π - arcsin (\frac{1}{4}) + 2 πn

解

2 x = arcsin (\frac{1}{4}) + 2 πn : x = \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 x = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 x = arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{1}{4} )}{2} + πn

x = \frac{arcsin ( \frac{1}{4} )}{2} + πn

解

2 x = π - arcsin (\frac{1}{4}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 x = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{1}{4}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

x = \frac{arcsin ( \frac{1}{4} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

合并所有解

x = πn, x = \frac{arcsin ( \frac{1}{4} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{4} )}{2} + πn

以小数形式表示解

x = πn, x = \frac{0.25268 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.25268 \dots}{2} + πn

作图

流行的例子

3sin(x)=sqrt(3)*sin(x)+3cos(x)+3sin(x)

3 sin (x) = 3 \cdot sin (x) + 3 cos (x) + 3 sin (x)

csc(θ)+1=0,(0,2pi)

csc (θ) + 1 = 0, (0, 2 π)

solvefor x,y^2=-2tan(2x)

so l v e f or x, y^{2} = - 2 tan (2 x)

2cos^2(a)-3cos(a)+1=0

2 cos^{2} (a) - 3 cos (a) + 1 = 0

sin(x)*cos(2x)=0.5-sin(2x)*cos(x)

sin (x) \cdot cos (2 x) = 0.5 - sin (2 x) \cdot cos (x)