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受欢迎的三角函数 >

2cos(2θ+10)=1

解答

2 cos (2 θ + 1 0^{\circ}) = 1

解答

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

+1

弧度

θ = \frac{5 π}{36} + πn, θ = \frac{29 π}{36} + πn

求解步骤

2 cos (2 θ + 1 0^{\circ}) = 1

两边除以

2

2 cos (2 θ + 1 0^{\circ}) = 1

两边除以

2

\frac{2 cos ( 2 θ + 1 0 ^{\circ} )}{2} = \frac{1}{2}

化简

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

cos (2 θ + 1 0^{\circ}) = \frac{1}{2}

的通解

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

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n, 2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n, 2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

解

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 2 5^{\circ}

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n

将

1 0^{\circ}

到右边

2 θ + 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n

两边减去

1 0^{\circ}

2 θ + 1 0^{\circ} - 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

化简

2 θ + 1 0^{\circ} - 1 0^{\circ} = 6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

化简

2 θ + 1 0^{\circ} - 1 0^{\circ} : 2 θ

2 θ + 1 0^{\circ} - 1 0^{\circ}

同类项相加:

1 0^{\circ} - 1 0^{\circ} = 0

= 2 θ

化简

6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 5 0^{\circ}

6 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

对同类项分组

= 36 0^{\circ} n + 6 0^{\circ} - 1 0^{\circ}

3, 18

的最小公倍数:

18

3, 18

最小公倍数 (LCM)

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

18

质因数分解:

2 \cdot 3 \cdot 3

18

18

除以

218 = 9 \cdot 2

= 2 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3 \cdot 3

将每个因子乘以它在

3

或

18

中出现的最多次数

= 3 \cdot 3 \cdot 2

数字相乘:

3 \cdot 3 \cdot 2 = 18

= 18

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

18

对于

6 0^{\circ} :

将分母和分子乘以

6

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

= 6 0^{\circ} - 1 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} 6 - 18 0 ^{\circ}}{18}

同类项相加:

108 0^{\circ} - 18 0^{\circ} = 90 0^{\circ}

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

2 θ = 36 0^{\circ} n + 5 0^{\circ}

2 θ = 36 0^{\circ} n + 5 0^{\circ}

2 θ = 36 0^{\circ} n + 5 0^{\circ}

两边除以

2

2 θ = 36 0^{\circ} n + 5 0^{\circ}

两边除以

2

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

化简

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{5 0 ^{\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{5 0 ^{\circ}}{2} : 18 0^{\circ} n + 2 5^{\circ}

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

\frac{5 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{90 0 ^{\circ}}{18 \cdot 2}

数字相乘:

18 \cdot 2 = 36

= 2 5^{\circ}

= 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}

解

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 14 5^{\circ}

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

将

1 0^{\circ}

到右边

2 θ + 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n

两边减去

1 0^{\circ}

2 θ + 1 0^{\circ} - 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

化简

2 θ + 1 0^{\circ} - 1 0^{\circ} = 30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

化简

2 θ + 1 0^{\circ} - 1 0^{\circ} : 2 θ

2 θ + 1 0^{\circ} - 1 0^{\circ}

同类项相加:

1 0^{\circ} - 1 0^{\circ} = 0

= 2 θ

化简

30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ} : 36 0^{\circ} n + 29 0^{\circ}

30 0^{\circ} + 36 0^{\circ} n - 1 0^{\circ}

对同类项分组

= 36 0^{\circ} n + 30 0^{\circ} - 1 0^{\circ}

3, 18

的最小公倍数:

18

3, 18

最小公倍数 (LCM)

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

18

质因数分解:

2 \cdot 3 \cdot 3

18

18

除以

218 = 9 \cdot 2

= 2 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3 \cdot 3

将每个因子乘以它在

3

或

18

中出现的最多次数

= 3 \cdot 3 \cdot 2

数字相乘:

3 \cdot 3 \cdot 2 = 18

= 18

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

18

对于

30 0^{\circ} :

将分母和分子乘以

6

30 0^{\circ} = \frac{90 0 ^{\circ} 6}{3 \cdot 6} = 30 0^{\circ}

= 30 0^{\circ} - 1 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{540 0 ^{\circ} - 18 0 ^{\circ}}{18}

同类项相加:

540 0^{\circ} - 18 0^{\circ} = 522 0^{\circ}

= 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

2 θ = 36 0^{\circ} n + 29 0^{\circ}

两边除以

2

2 θ = 36 0^{\circ} n + 29 0^{\circ}

两边除以

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\circ}}{2}

化简

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\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{29 0 ^{\circ}}{2} : 18 0^{\circ} n + 14 5^{\circ}

\frac{36 0 ^{\circ} n}{2} + \frac{29 0 ^{\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{29 0 ^{\circ}}{2} = 14 5^{\circ}

\frac{29 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{522 0 ^{\circ}}{18 \cdot 2}

数字相乘:

18 \cdot 2 = 36

= 14 5^{\circ}

= 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 14 5^{\circ}

θ = 18 0^{\circ} n + 2 5^{\circ}, θ = 18 0^{\circ} n + 14 5^{\circ}

作图

流行的例子

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cot (β) = \frac{- 3}{3}

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sin (θ) = \frac{15}{23}

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