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受欢迎的三角函数 >

2sin(e^{t/4})+1=0

解答

2 sin (e^{\frac{t}{4}}) + 1 = 0

解答

t = 4 ln (\frac{7 π}{6} + 2 πn), t = 4 ln (\frac{11 π}{6} + 2 πn)

+1

度数

t = 0^{\circ} + 526.52744 \dots^{\circ} n, t = 0^{\circ} + 570.31408 \dots^{\circ} n

求解步骤

2 sin (e^{\frac{t}{4}}) + 1 = 0

将

1

到右边

2 sin (e^{\frac{t}{4}}) + 1 = 0

两边减去

1

2 sin (e^{\frac{t}{4}}) + 1 - 1 = 0 - 1

化简

2 sin (e^{\frac{t}{4}}) = - 1

2 sin (e^{\frac{t}{4}}) = - 1

两边除以

2

2 sin (e^{\frac{t}{4}}) = - 1

两边除以

2

\frac{2 sin ( e ^{\frac{t}{4}} )}{2} = \frac{- 1}{2}

化简

sin (e^{\frac{t}{4}}) = - \frac{1}{2}

sin (e^{\frac{t}{4}}) = - \frac{1}{2}

sin (e^{\frac{t}{4}}) = - \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn, e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn, e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

解

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn : t = 4 ln (\frac{7 π}{6} + 2 πn)

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn

在两边乘以

6

e^{\frac{t}{4}} \cdot 6 = \frac{7 π}{6} \cdot 6 + 2 πn \cdot 6

化简

e^{\frac{t}{4}} \cdot 6 = 7 π + 12 πn

两边除以

6

e^{\frac{t}{4}} \cdot 6 = 7 π + 12 πn

两边除以

6

\frac{e ^{\frac{t}{4}} \cdot 6}{6} = \frac{7 π}{6} + \frac{12 πn}{6}

化简

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn

使用指数运算法则

e^{\frac{t}{4}} = \frac{7 π}{6} + 2 πn

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{\frac{t}{4}}) = ln (\frac{7 π}{6} + 2 πn)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{\frac{t}{4}}) = \frac{t}{4}

\frac{t}{4} = ln (\frac{7 π}{6} + 2 πn)

\frac{t}{4} = ln (\frac{7 π}{6} + 2 πn)

解

\frac{t}{4} = ln (\frac{7 π}{6} + 2 πn) : t = 4 ln (\frac{7 π}{6} + 2 πn)

\frac{t}{4} = ln (\frac{7 π}{6} + 2 πn)

在两边乘以

4

\frac{t}{4} = ln (\frac{7 π}{6} + 2 πn)

在两边乘以

4

\frac{4 t}{4} = 4 ln (\frac{7 π}{6} + 2 πn)

化简

t = 4 ln (\frac{7 π}{6} + 2 πn)

t = 4 ln (\frac{7 π}{6} + 2 πn)

t = 4 ln (\frac{7 π}{6} + 2 πn)

解

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn : t = 4 ln (\frac{11 π}{6} + 2 πn)

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

在两边乘以

6

e^{\frac{t}{4}} \cdot 6 = \frac{11 π}{6} \cdot 6 + 2 πn \cdot 6

化简

e^{\frac{t}{4}} \cdot 6 = 11 π + 12 πn

两边除以

6

e^{\frac{t}{4}} \cdot 6 = 11 π + 12 πn

两边除以

6

\frac{e ^{\frac{t}{4}} \cdot 6}{6} = \frac{11 π}{6} + \frac{12 πn}{6}

化简

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

使用指数运算法则

e^{\frac{t}{4}} = \frac{11 π}{6} + 2 πn

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{\frac{t}{4}}) = ln (\frac{11 π}{6} + 2 πn)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{\frac{t}{4}}) = \frac{t}{4}

\frac{t}{4} = ln (\frac{11 π}{6} + 2 πn)

\frac{t}{4} = ln (\frac{11 π}{6} + 2 πn)

解

\frac{t}{4} = ln (\frac{11 π}{6} + 2 πn) : t = 4 ln (\frac{11 π}{6} + 2 πn)

\frac{t}{4} = ln (\frac{11 π}{6} + 2 πn)

在两边乘以

4

\frac{t}{4} = ln (\frac{11 π}{6} + 2 πn)

在两边乘以

4

\frac{4 t}{4} = 4 ln (\frac{11 π}{6} + 2 πn)

化简

t = 4 ln (\frac{11 π}{6} + 2 πn)

t = 4 ln (\frac{11 π}{6} + 2 πn)

t = 4 ln (\frac{11 π}{6} + 2 πn)

t = 4 ln (\frac{7 π}{6} + 2 πn), t = 4 ln (\frac{11 π}{6} + 2 πn)

作图

流行的例子

3sin^2(x)-2sin(x)=0

3 sin^{2} (x) - 2 sin (x) = 0

sin (u) = - \frac{2}{5}

2sin(3x)cos(2x)-cos(2x)=0

2 sin (3 x) cos (2 x) - cos (2 x) = 0

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

4 sin^{2} (x) + 2 sin (x) - 2 = 0

solvefor x,cos(xy)=0

so l v e f or x, cos (x y) = 0