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受欢迎的三角函数 >

3cos(2x+1)=2

解答

3 cos (2 x + 1) = 2

解答

x = πn - \frac{1}{2} + \frac{0.84106 \dots}{2}, x = π + πn - \frac{1}{2} - \frac{0.84106 \dots}{2}

+1

度数

x = - 4.55304 \dots^{\circ} + 18 0^{\circ} n, x = 127.25726 \dots^{\circ} + 18 0^{\circ} n

求解步骤

3 cos (2 x + 1) = 2

两边除以

3

3 cos (2 x + 1) = 2

两边除以

3

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

化简

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

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

使用反三角函数性质

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

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

的通解

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

2 x + 1 = arccos (\frac{2}{3}) + 2 πn, 2 x + 1 = 2 π - arccos (\frac{2}{3}) + 2 πn

2 x + 1 = arccos (\frac{2}{3}) + 2 πn, 2 x + 1 = 2 π - arccos (\frac{2}{3}) + 2 πn

解

2 x + 1 = arccos (\frac{2}{3}) + 2 πn : x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

2 x + 1 = arccos (\frac{2}{3}) + 2 πn

将

1

到右边

2 x + 1 = arccos (\frac{2}{3}) + 2 πn

两边减去

1

2 x + 1 - 1 = arccos (\frac{2}{3}) + 2 πn - 1

化简

2 x = arccos (\frac{2}{3}) + 2 πn - 1

2 x = arccos (\frac{2}{3}) + 2 πn - 1

两边除以

2

2 x = arccos (\frac{2}{3}) + 2 πn - 1

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2}

化简

\frac{2 x}{2} = \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2} : πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

\frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2}

对同类项分组

= - \frac{1}{2} + \frac{2 πn}{2} + \frac{arccos ( \frac{2}{3} )}{2}

数字相除:

\frac{2}{2} = 1

= - \frac{1}{2} + πn + \frac{arccos ( \frac{2}{3} )}{2}

对同类项分组

= πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}

解

2 x + 1 = 2 π - arccos (\frac{2}{3}) + 2 πn : x = π + πn - \frac{1}{2} - \frac{arccos ( \frac{2}{3} )}{2}

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

将

1

到右边

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

两边减去

1

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

化简

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

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

两边除以

2

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

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 π}{2} - \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2} : π + πn - \frac{1}{2} - \frac{arccos ( \frac{2}{3} )}{2}

\frac{2 π}{2} - \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2} - \frac{1}{2}

对同类项分组

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

数字相除:

\frac{2}{2} = 1

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

对同类项分组

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

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

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

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

x = πn - \frac{1}{2} + \frac{arccos ( \frac{2}{3} )}{2}, x = π + πn - \frac{1}{2} - \frac{arccos ( \frac{2}{3} )}{2}

以小数形式表示解

x = πn - \frac{1}{2} + \frac{0.84106 \dots}{2}, x = π + πn - \frac{1}{2} - \frac{0.84106 \dots}{2}

作图

流行的例子

sin(t)= 3/7 ,cos(t)

sin (t) = \frac{3}{7}, cos (t)

tan (x) = \frac{16}{12}

sin(4x+10)=sin(90)

sin (4 x + 10) = sin (9 0^{\circ})

solvefor y,-1/2 cot(y)=(t^2)/(+2)+C

so l v e f or y, - \frac{1}{2} cot (y) = \frac{t ^{2}}{+ 2} + C

sqrt(3)sec(2θ)+2=0

3 sec (2 θ) + 2 = 0