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受欢迎的三角函数 >

6cos(a)+sqrt(18)=0

解答

6 cos (a) + 18 = 0

解答

a = \frac{3 π}{4} + 2 πn, a = \frac{5 π}{4} + 2 πn

+1

度数

a = 13 5^{\circ} + 36 0^{\circ} n, a = 22 5^{\circ} + 36 0^{\circ} n

求解步骤

6 cos (a) + 18 = 0

18 = 32

18

18

质因数分解:

2 \cdot 3^{2}

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

= 2 \cdot 3^{2}

= 3^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 32

6 cos (a) + 32 = 0

将

32

到右边

6 cos (a) + 32 = 0

两边减去

32

6 cos (a) + 32 - 32 = 0 - 32

化简

6 cos (a) = - 32

6 cos (a) = - 32

两边除以

6

6 cos (a) = - 32

两边除以

6

\frac{6 cos ( a )}{6} = \frac{- 3 2}{6}

化简

\frac{6 cos ( a )}{6} = \frac{- 3 2}{6}

化简

\frac{6 cos ( a )}{6} : cos (a)

\frac{6 cos ( a )}{6}

数字相除:

\frac{6}{6} = 1

= cos (a)

化简

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

\frac{- 3 2}{6}

使用分式法则:

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

= - \frac{3 2}{6}

约分:

3

= - \frac{2}{2}

cos (a) = - \frac{2}{2}

cos (a) = - \frac{2}{2}

cos (a) = - \frac{2}{2}

cos (a) = - \frac{2}{2}

的通解

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

a = \frac{3 π}{4} + 2 πn, a = \frac{5 π}{4} + 2 πn

a = \frac{3 π}{4} + 2 πn, a = \frac{5 π}{4} + 2 πn

作图

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