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受欢迎的三角函数 >

3cos(2θ)-5cos(θ)-1=0

解答

3 cos (2 θ) - 5 cos (θ) - 1 = 0

解答

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

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 θ) - 5 cos (θ) - 1 = 0

使用三角恒等式改写

- 1 + 3 cos (2 θ) - 5 cos (θ)

使用倍角公式:

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

= - 1 + 3 (2 cos^{2} (θ) - 1) - 5 cos (θ)

化简

- 1 + 3 (2 cos^{2} (θ) - 1) - 5 cos (θ) : 6 cos^{2} (θ) - 5 cos (θ) - 4

- 1 + 3 (2 cos^{2} (θ) - 1) - 5 cos (θ)

乘开

3 (2 cos^{2} (θ) - 1) : 6 cos^{2} (θ) - 3

3 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (θ), c = 1

= 3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

化简

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1 : 6 cos^{2} (θ) - 3

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

数字相乘:

3 \cdot 2 = 6

= 6 cos^{2} (θ) - 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 6 cos^{2} (θ) - 3

= 6 cos^{2} (θ) - 3

= - 1 + 6 cos^{2} (θ) - 3 - 5 cos (θ)

化简

- 1 + 6 cos^{2} (θ) - 3 - 5 cos (θ) : 6 cos^{2} (θ) - 5 cos (θ) - 4

- 1 + 6 cos^{2} (θ) - 3 - 5 cos (θ)

对同类项分组

= 6 cos^{2} (θ) - 5 cos (θ) - 1 - 3

数字相减:

- 1 - 3 = - 4

= 6 cos^{2} (θ) - 5 cos (θ) - 4

= 6 cos^{2} (θ) - 5 cos (θ) - 4

= 6 cos^{2} (θ) - 5 cos (θ) - 4

- 4 - 5 cos (θ) + 6 cos^{2} (θ) = 0

用替代法求解

- 4 - 5 cos (θ) + 6 cos^{2} (θ) = 0

令

: cos (θ) = u

- 4 - 5 u + 6 u^{2} = 0

- 4 - 5 u + 6 u^{2} = 0 : u = \frac{4}{3}, u = - \frac{1}{2}

- 4 - 5 u + 6 u^{2} = 0

改写成标准形式

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

6 u^{2} - 5 u - 4 = 0

使用求根公式求解

6 u^{2} - 5 u - 4 = 0

二次方程求根公式:

若

a = 6, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 4 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 4 )}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 (- 4) = 11

(- 5)^{2} - 4 \cdot 6 (- 4)

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 6 \cdot 4

数字相乘:

4 \cdot 6 \cdot 4 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

数字相加:

25 + 96 = 121

= 121

因式分解数字:

121 = 1 1^{2}

= 1 1^{2}

使用根式运算法则:

n a^{n} = a

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 6}

u = \frac{- ( - 5 ) + 11}{2 \cdot 6} : \frac{4}{3}

\frac{- ( - 5 ) + 11}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 + 11}{2 \cdot 6}

数字相加:

5 + 11 = 16

= \frac{16}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{16}{12}

约分:

4

= \frac{4}{3}

u = \frac{- ( - 5 ) - 11}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 5 ) - 11}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 - 11}{2 \cdot 6}

数字相减:

5 - 11 = - 6

= \frac{- 6}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

二次方程组的解是:

u = \frac{4}{3}, u = - \frac{1}{2}

u = cos (θ)

代回

cos (θ) = \frac{4}{3}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{4}{3}, cos (θ) = - \frac{1}{2}

cos (θ) = \frac{4}{3} :

无解

cos (θ) = \frac{4}{3}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

cos (θ) = - \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}

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

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

合并所有解

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

作图

流行的例子

csc (x) = \frac{21}{5}

sqrt(3)cot(2x)+3=0

3 cot (2 x) + 3 = 0

\frac{1}{cos ( θ )} = 2

cos(θ)= 7/25 ,(3pi)/2 <θ<2pi

cos (θ) = \frac{7}{25}, \frac{3 π}{2} < θ < 2 π

solvefor x,cos(x)=-1

so l v e f or x, cos (x) = - 1