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受欢迎的三角函数 >

-8sin(2θ+45)=-4,0<= x<360

解答

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

解答

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

+1

弧度

θ = \frac{7 π}{24}, θ = \frac{23 π}{24}, θ = \frac{31 π}{24}, θ = \frac{47 π}{24}

求解步骤

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

两边除以

- 8

- 8 sin (2 θ + 4 5^{\circ}) = - 4

两边除以

- 8

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} = \frac{- 4}{- 8}

化简

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} = \frac{- 4}{- 8}

化简

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8} : sin (2 θ + 4 5^{\circ})

\frac{- 8 sin ( 2 θ + 4 5 ^{\circ} )}{- 8}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{8 sin ( 2 θ + 4 5 ^{\circ} )}{8}

数字相除:

\frac{8}{8} = 1

= sin (2 θ + 4 5^{\circ})

化简

\frac{- 4}{- 8} : \frac{1}{2}

\frac{- 4}{- 8}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{8}

约分:

4

= \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

解

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n - 7. 5^{\circ}

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

将

4 5^{\circ}

到右边

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n

两边减去

4 5^{\circ}

2 θ + 4 5^{\circ} - 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

化简

2 θ + 4 5^{\circ} - 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

化简

2 θ + 4 5^{\circ} - 4 5^{\circ} : 2 θ

2 θ + 4 5^{\circ} - 4 5^{\circ}

同类项相加:

4 5^{\circ} - 4 5^{\circ} = 0

= 2 θ

化简

3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ} : 36 0^{\circ} n - 1 5^{\circ}

3 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

对同类项分组

= 36 0^{\circ} n + 3 0^{\circ} - 4 5^{\circ}

6, 4

的最小公倍数:

12

6, 4

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

6

或

4

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

3 0^{\circ} :

将分母和分子乘以

2

3 0^{\circ} = \frac{18 0 ^{\circ} 2}{6 \cdot 2} = 3 0^{\circ}

对于

4 5^{\circ} :

将分母和分子乘以

3

4 5^{\circ} = \frac{18 0 ^{\circ} 3}{4 \cdot 3} = 4 5^{\circ}

= 3 0^{\circ} - 4 5^{\circ}

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{18 0 ^{\circ} 2 - 18 0 ^{\circ} 3}{12}

同类项相加:

36 0^{\circ} - 54 0^{\circ} = - 18 0^{\circ}

= \frac{- 18 0 ^{\circ}}{12}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

2 θ = 36 0^{\circ} n - 1 5^{\circ}

两边除以

2

2 θ = 36 0^{\circ} n - 1 5^{\circ}

两边除以

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

化简

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2} : 18 0^{\circ} n - 7. 5^{\circ}

\frac{36 0 ^{\circ} n}{2} - \frac{1 5 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{1 5 ^{\circ}}{2} = 7. 5^{\circ}

\frac{1 5 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= 7. 5^{\circ}

= 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}

解

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 52. 5^{\circ}

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

将

4 5^{\circ}

到右边

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

两边减去

4 5^{\circ}

2 θ + 4 5^{\circ} - 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

化简

2 θ + 4 5^{\circ} - 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

化简

2 θ + 4 5^{\circ} - 4 5^{\circ} : 2 θ

2 θ + 4 5^{\circ} - 4 5^{\circ}

同类项相加:

4 5^{\circ} - 4 5^{\circ} = 0

= 2 θ

化简

15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ} : 36 0^{\circ} n + 10 5^{\circ}

15 0^{\circ} + 36 0^{\circ} n - 4 5^{\circ}

对同类项分组

= 36 0^{\circ} n - 4 5^{\circ} + 15 0^{\circ}

4, 6

的最小公倍数:

12

4, 6

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

4

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

4 5^{\circ} :

将分母和分子乘以

3

4 5^{\circ} = \frac{18 0 ^{\circ} 3}{4 \cdot 3} = 4 5^{\circ}

对于

15 0^{\circ} :

将分母和分子乘以

2

15 0^{\circ} = \frac{90 0 ^{\circ} 2}{6 \cdot 2} = 15 0^{\circ}

= - 4 5^{\circ} + 15 0^{\circ}

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{- 18 0 ^{\circ} 3 + 180 0 ^{\circ}}{12}

同类项相加:

- 54 0^{\circ} + 180 0^{\circ} = 126 0^{\circ}

= 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

2 θ = 36 0^{\circ} n + 10 5^{\circ}

两边除以

2

2 θ = 36 0^{\circ} n + 10 5^{\circ}

两边除以

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

化简

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2} : 18 0^{\circ} n + 52. 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{10 5 ^{\circ}}{2}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数字相除:

\frac{2}{2} = 1

= 18 0^{\circ} n

\frac{10 5 ^{\circ}}{2} = 52. 5^{\circ}

\frac{10 5 ^{\circ}}{2}

使用分式法则:

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

= \frac{126 0 ^{\circ}}{12 \cdot 2}

数字相乘:

12 \cdot 2 = 24

= 52. 5^{\circ}

= 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}, θ = 18 0^{\circ} n + 52. 5^{\circ}

在

0 \leq x < 36 0^{\circ}

范围内的解

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

作图

流行的例子

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sin (b) = \frac{10.21 ( sin ( 61.3 6 ^{\circ} ) )}{11.47}

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sin(θ)=-2/3 ,180<θ<270

sin (θ) = - \frac{2}{3}, 18 0^{\circ} < θ < 27 0^{\circ}

solvefor x,Rf=e^{cos(2x)}

so l v e f or x, R f = e^{c o s (2 x)}