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受欢迎的三角函数 >

8+4cos(2x)=12cos(x)

解答

8 + 4 cos (2 x) = 12 cos (x)

解答

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

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

8 + 4 cos (2 x) = 12 cos (x)

两边减去

12 cos (x)

8 + 4 cos (2 x) - 12 cos (x) = 0

使用三角恒等式改写

8 - 12 cos (x) + 4 cos (2 x)

使用倍角公式:

cos (2 x) = 2 cos^{2} (x) - 1

= 8 - 12 cos (x) + 4 (2 cos^{2} (x) - 1)

化简

8 - 12 cos (x) + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 12 cos (x) + 4

8 - 12 cos (x) + 4 (2 cos^{2} (x) - 1)

乘开

4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 4

4 (2 cos^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

= 4 \cdot 2 cos^{2} (x) - 4 \cdot 1

化简

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

4 \cdot 2 cos^{2} (x) - 4 \cdot 1

数字相乘:

4 \cdot 2 = 8

= 8 cos^{2} (x) - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= 8 - 12 cos (x) + 8 cos^{2} (x) - 4

化简

8 - 12 cos (x) + 8 cos^{2} (x) - 4 : 8 cos^{2} (x) - 12 cos (x) + 4

8 - 12 cos (x) + 8 cos^{2} (x) - 4

对同类项分组

= - 12 cos (x) + 8 cos^{2} (x) + 8 - 4

数字相加/相减:

8 - 4 = 4

= 8 cos^{2} (x) - 12 cos (x) + 4

= 8 cos^{2} (x) - 12 cos (x) + 4

= 8 cos^{2} (x) - 12 cos (x) + 4

4 - 12 cos (x) + 8 cos^{2} (x) = 0

用替代法求解

4 - 12 cos (x) + 8 cos^{2} (x) = 0

令

: cos (x) = u

4 - 12 u + 8 u^{2} = 0

4 - 12 u + 8 u^{2} = 0 : u = 1, u = \frac{1}{2}

4 - 12 u + 8 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

8 u^{2} - 12 u + 4 = 0

使用求根公式求解

8 u^{2} - 12 u + 4 = 0

二次方程求根公式:

若

a = 8, b = - 12, c = 4

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 8 \cdot 4}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 8 \cdot 4}{2 \cdot 8}

(- 12)^{2} - 4 \cdot 8 \cdot 4 = 4

(- 12)^{2} - 4 \cdot 8 \cdot 4

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 8 \cdot 4

数字相乘:

4 \cdot 8 \cdot 4 = 128

= 1 2^{2} - 128

1 2^{2} = 144

= 144 - 128

数字相减:

144 - 128 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 12 ) \pm 4}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 8}, u_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 8}

u = \frac{- ( - 12 ) + 4}{2 \cdot 8} : 1

\frac{- ( - 12 ) + 4}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{12 + 4}{2 \cdot 8}

数字相加:

12 + 4 = 16

= \frac{16}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{16}{16}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 12 ) - 4}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 12 ) - 4}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{12 - 4}{2 \cdot 8}

数字相减:

12 - 4 = 8

= \frac{8}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

二次方程组的解是:

u = 1, u = \frac{1}{2}

u = cos (x)

代回

cos (x) = 1, cos (x) = \frac{1}{2}

cos (x) = 1, cos (x) = \frac{1}{2}

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{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}

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

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

合并所有解

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

作图

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