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受欢迎的三角函数 >

2=4cos(3x+1)

解答

2 = 4 cos (3 x + 1)

解答

x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}, x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

+1

度数

x = 0.90140 \dots^{\circ} + 12 0^{\circ} n, x = 80.90140 \dots^{\circ} + 12 0^{\circ} n

求解步骤

2 = 4 cos (3 x + 1)

交换两边

4 cos (3 x + 1) = 2

两边除以

4

4 cos (3 x + 1) = 2

两边除以

4

\frac{4 cos ( 3 x + 1 )}{4} = \frac{2}{4}

化简

cos (3 x + 1) = \frac{1}{2}

cos (3 x + 1) = \frac{1}{2}

cos (3 x + 1) = \frac{1}{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}

3 x + 1 = \frac{π}{3} + 2 πn, 3 x + 1 = \frac{5 π}{3} + 2 πn

3 x + 1 = \frac{π}{3} + 2 πn, 3 x + 1 = \frac{5 π}{3} + 2 πn

解

3 x + 1 = \frac{π}{3} + 2 πn : x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

3 x + 1 = \frac{π}{3} + 2 πn

将

1

到右边

3 x + 1 = \frac{π}{3} + 2 πn

两边减去

1

3 x + 1 - 1 = \frac{π}{3} + 2 πn - 1

化简

3 x = \frac{π}{3} + 2 πn - 1

3 x = \frac{π}{3} + 2 πn - 1

两边除以

3

3 x = \frac{π}{3} + 2 πn - 1

两边除以

3

\frac{3 x}{3} = \frac{\frac{π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

化简

\frac{3 x}{3} = \frac{\frac{π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3} : - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

对同类项分组

= - \frac{1}{3} + \frac{2 πn}{3} + \frac{\frac{π}{3}}{3}

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

使用分式法则:

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

= \frac{π}{3 \cdot 3}

数字相乘:

3 \cdot 3 = 9

= \frac{π}{9}

= - \frac{1}{3} + \frac{2 πn}{3} + \frac{π}{9}

对同类项分组

= - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}

解

3 x + 1 = \frac{5 π}{3} + 2 πn : x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

3 x + 1 = \frac{5 π}{3} + 2 πn

将

1

到右边

3 x + 1 = \frac{5 π}{3} + 2 πn

两边减去

1

3 x + 1 - 1 = \frac{5 π}{3} + 2 πn - 1

化简

3 x = \frac{5 π}{3} + 2 πn - 1

3 x = \frac{5 π}{3} + 2 πn - 1

两边除以

3

3 x = \frac{5 π}{3} + 2 πn - 1

两边除以

3

\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

化简

\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3} : - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} - \frac{1}{3}

对同类项分组

= - \frac{1}{3} + \frac{2 πn}{3} + \frac{\frac{5 π}{3}}{3}

\frac{\frac{5 π}{3}}{3} = \frac{5 π}{9}

\frac{\frac{5 π}{3}}{3}

使用分式法则:

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

= \frac{5 π}{3 \cdot 3}

数字相乘:

3 \cdot 3 = 9

= \frac{5 π}{9}

= - \frac{1}{3} + \frac{2 πn}{3} + \frac{5 π}{9}

对同类项分组

= - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

x = - \frac{1}{3} + \frac{π}{9} + \frac{2 πn}{3}, x = - \frac{1}{3} + \frac{5 π}{9} + \frac{2 πn}{3}

作图

流行的例子

tan (x) = \frac{13}{6}

3 sinh (2 x) = 5

sin(x/(25.4))=35

sin (\frac{x}{25.4}) = 35

2/3 =(sin(x))/(sin(135))

\frac{2}{3} = \frac{sin ( x )}{sin ( 13 5 ^{\circ} )}

sin(x)+2(cos(x))^2=1

sin (x) + 2 (cos (x))^{2} = 1