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受欢迎的三角函数 >

pi/3 =3+2sin(2t)

解答

\frac{π}{3} = 3 + 2 sin (2 t)

解答

t = \frac{- 1.35311 \dots}{2} + πn, t = \frac{π}{2} + \frac{1.35311 \dots}{2} + πn

+1

度数

t = - 38.76394 \dots^{\circ} + 18 0^{\circ} n, t = 128.76394 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{π}{3} = 3 + 2 sin (2 t)

交换两边

3 + 2 sin (2 t) = \frac{π}{3}

将

3

到右边

3 + 2 sin (2 t) = \frac{π}{3}

两边减去

3

3 + 2 sin (2 t) - 3 = \frac{π}{3} - 3

化简

2 sin (2 t) = \frac{π}{3} - 3

2 sin (2 t) = \frac{π}{3} - 3

两边除以

2

2 sin (2 t) = \frac{π}{3} - 3

两边除以

2

\frac{2 sin ( 2 t )}{2} = \frac{\frac{π}{3}}{2} - \frac{3}{2}

化简

\frac{2 sin ( 2 t )}{2} = \frac{\frac{π}{3}}{2} - \frac{3}{2}

化简

\frac{2 sin ( 2 t )}{2} : sin (2 t)

\frac{2 sin ( 2 t )}{2}

数字相除:

\frac{2}{2} = 1

= sin (2 t)

化简

\frac{\frac{π}{3}}{2} - \frac{3}{2} : \frac{π - 9}{6}

\frac{\frac{π}{3}}{2} - \frac{3}{2}

使用法则

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

= \frac{\frac{π}{3} - 3}{2}

化简

\frac{π}{3} - 3 : \frac{π - 9}{3}

\frac{π}{3} - 3

将项转换为分式:

3 = \frac{3 \cdot 3}{3}

= \frac{π}{3} - \frac{3 \cdot 3}{3}

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{π - 3 \cdot 3}{3}

数字相乘:

3 \cdot 3 = 9

= \frac{π - 9}{3}

= \frac{\frac{π - 9}{3}}{2}

使用分式法则:

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

= \frac{π - 9}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π - 9}{6}

sin (2 t) = \frac{π - 9}{6}

sin (2 t) = \frac{π - 9}{6}

sin (2 t) = \frac{π - 9}{6}

使用反三角函数性质

sin (2 t) = \frac{π - 9}{6}

sin (2 t) = \frac{π - 9}{6}

的通解

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

2 t = arcsin (\frac{π - 9}{6}) + 2 πn, 2 t = π + arcsin (- \frac{π - 9}{6}) + 2 πn

2 t = arcsin (\frac{π - 9}{6}) + 2 πn, 2 t = π + arcsin (- \frac{π - 9}{6}) + 2 πn

解

2 t = arcsin (\frac{π - 9}{6}) + 2 πn : t = \frac{arcsin ( \frac{π - 9}{6} )}{2} + πn

2 t = arcsin (\frac{π - 9}{6}) + 2 πn

两边除以

2

2 t = arcsin (\frac{π - 9}{6}) + 2 πn

两边除以

2

\frac{2 t}{2} = \frac{arcsin ( \frac{π - 9}{6} )}{2} + \frac{2 πn}{2}

化简

t = \frac{arcsin ( \frac{π - 9}{6} )}{2} + πn

t = \frac{arcsin ( \frac{π - 9}{6} )}{2} + πn

解

2 t = π + arcsin (- \frac{π - 9}{6}) + 2 πn : t = \frac{π}{2} + \frac{arcsin ( \frac{- π + 9}{6} )}{2} + πn

2 t = π + arcsin (- \frac{π - 9}{6}) + 2 πn

- \frac{π - 9}{6} = - \frac{- ( - π + 9 )}{6} = \frac{- π + 9}{6}

2 t = π + arcsin (\frac{- π + 9}{6}) + 2 πn

两边除以

2

2 t = π + arcsin (\frac{- π + 9}{6}) + 2 πn

两边除以

2

\frac{2 t}{2} = \frac{π}{2} + \frac{arcsin ( \frac{- π + 9}{6} )}{2} + \frac{2 πn}{2}

化简

t = \frac{π}{2} + \frac{arcsin ( \frac{- π + 9}{6} )}{2} + πn

t = \frac{π}{2} + \frac{arcsin ( \frac{- π + 9}{6} )}{2} + πn

t = \frac{arcsin ( \frac{π - 9}{6} )}{2} + πn, t = \frac{π}{2} + \frac{arcsin ( \frac{- π + 9}{6} )}{2} + πn

以小数形式表示解

t = \frac{- 1.35311 \dots}{2} + πn, t = \frac{π}{2} + \frac{1.35311 \dots}{2} + πn

作图

流行的例子

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