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受欢迎的三角函数 >

h=3sin(2x-1)+4

解答

h = 3 sin (2 x - 1) + 4

解答

x = \frac{arcsin ( \frac{h - 4}{3} )}{2} + πn + \frac{1}{2}, x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

求解步骤

h = 3 sin (2 x - 1) + 4

交换两边

3 sin (2 x - 1) + 4 = h

将

4

到右边

3 sin (2 x - 1) + 4 = h

两边减去

4

3 sin (2 x - 1) + 4 - 4 = h - 4

化简

3 sin (2 x - 1) = h - 4

3 sin (2 x - 1) = h - 4

两边除以

3

3 sin (2 x - 1) = h - 4

两边除以

3

\frac{3 sin ( 2 x - 1 )}{3} = \frac{h}{3} - \frac{4}{3}

化简

\frac{3 sin ( 2 x - 1 )}{3} = \frac{h}{3} - \frac{4}{3}

化简

\frac{3 sin ( 2 x - 1 )}{3} : sin (2 x - 1)

\frac{3 sin ( 2 x - 1 )}{3}

数字相除:

\frac{3}{3} = 1

= sin (2 x - 1)

化简

\frac{h}{3} - \frac{4}{3} : \frac{h - 4}{3}

\frac{h}{3} - \frac{4}{3}

使用法则

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

= \frac{h - 4}{3}

sin (2 x - 1) = \frac{h - 4}{3}

sin (2 x - 1) = \frac{h - 4}{3}

sin (2 x - 1) = \frac{h - 4}{3}

使用反三角函数性质

sin (2 x - 1) = \frac{h - 4}{3}

sin (2 x - 1) = \frac{h - 4}{3}

的通解

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

2 x - 1 = arcsin (\frac{h - 4}{3}) + 2 πn, 2 x - 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn

2 x - 1 = arcsin (\frac{h - 4}{3}) + 2 πn, 2 x - 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn

解

2 x - 1 = arcsin (\frac{h - 4}{3}) + 2 πn : x = \frac{arcsin ( \frac{h - 4}{3} )}{2} + πn + \frac{1}{2}

2 x - 1 = arcsin (\frac{h - 4}{3}) + 2 πn

将

1

到右边

2 x - 1 = arcsin (\frac{h - 4}{3}) + 2 πn

两边加上

1

2 x - 1 + 1 = arcsin (\frac{h - 4}{3}) + 2 πn + 1

化简

2 x = arcsin (\frac{h - 4}{3}) + 2 πn + 1

2 x = arcsin (\frac{h - 4}{3}) + 2 πn + 1

两边除以

2

2 x = arcsin (\frac{h - 4}{3}) + 2 πn + 1

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{h - 4}{3} )}{2} + \frac{2 πn}{2} + \frac{1}{2}

化简

x = \frac{arcsin ( \frac{h - 4}{3} )}{2} + πn + \frac{1}{2}

x = \frac{arcsin ( \frac{h - 4}{3} )}{2} + πn + \frac{1}{2}

解

2 x - 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn : x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

2 x - 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn

将

1

到右边

2 x - 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn

两边加上

1

2 x - 1 + 1 = π + arcsin (- \frac{h - 4}{3}) + 2 πn + 1

化简

2 x = π + arcsin (- \frac{h - 4}{3}) + 2 πn + 1

2 x = π + arcsin (- \frac{h - 4}{3}) + 2 πn + 1

两边除以

2

2 x = π + arcsin (- \frac{h - 4}{3}) + 2 πn + 1

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( - \frac{h - 4}{3} )}{2} + \frac{2 πn}{2} + \frac{1}{2}

化简

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( - \frac{h - 4}{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{π}{2} + \frac{arcsin ( - \frac{h - 4}{3} )}{2} + \frac{2 πn}{2} + \frac{1}{2} : \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

\frac{π}{2} + \frac{arcsin ( - \frac{h - 4}{3} )}{2} + \frac{2 πn}{2} + \frac{1}{2}

对同类项分组

= \frac{π}{2} + \frac{1}{2} + \frac{2 πn}{2} + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

数字相除:

\frac{2}{2} = 1

= \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

x = \frac{arcsin ( \frac{h - 4}{3} )}{2} + πn + \frac{1}{2}, x = \frac{π}{2} + \frac{1}{2} + πn + \frac{arcsin ( - \frac{h - 4}{3} )}{2}

作图

流行的例子

sin^2(x)+2=3sin(x)

sin^{2} (x) + 2 = 3 sin (x)

sin(x+pi/3)= 1/2 cos(x-pi/6)

sin (x + \frac{π}{3}) = \frac{1}{2} cos (x - \frac{π}{6})

(2sec^2(x)-1)/(sec^2(x))=sec^2(x)

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9(sin(x)-0.6pi)+8=0

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solvefor t,v=100cos(50t+20)v

so l v e f or t, v = 100 cos (50 t + 2 0^{\circ}) v