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受欢迎的三角函数 >

cos(2θ)+5cos(θ)+3=0

解答

cos (2 θ) + 5 cos (θ) + 3 = 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

cos (2 θ) + 5 cos (θ) + 3 = 0

使用三角恒等式改写

3 + cos (2 θ) + 5 cos (θ)

使用倍角公式:

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

= 3 + 2 cos^{2} (θ) - 1 + 5 cos (θ)

化简

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

3 + 2 cos^{2} (θ) - 1 + 5 cos (θ)

对同类项分组

= 2 cos^{2} (θ) + 5 cos (θ) + 3 - 1

数字相加/相减:

3 - 1 = 2

= 2 cos^{2} (θ) + 5 cos (θ) + 2

= 2 cos^{2} (θ) + 5 cos (θ) + 2

2 + 2 cos^{2} (θ) + 5 cos (θ) = 0

用替代法求解

2 + 2 cos^{2} (θ) + 5 cos (θ) = 0

令

: cos (θ) = u

2 + 2 u^{2} + 5 u = 0

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

2 + 2 u^{2} + 5 u = 0

改写成标准形式

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

2 u^{2} + 5 u + 2 = 0

使用求根公式求解

2 u^{2} + 5 u + 2 = 0

二次方程求根公式:

若

a = 2, b = 5, c = 2

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

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

5^{2} - 4 \cdot 2 \cdot 2 = 3

5^{2} - 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 5 + 3}{2 \cdot 2}, u_{2} = \frac{- 5 - 3}{2 \cdot 2}

u = \frac{- 5 + 3}{2 \cdot 2} : - \frac{1}{2}

\frac{- 5 + 3}{2 \cdot 2}

数字相加/相减:

- 5 + 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

u = \frac{- 5 - 3}{2 \cdot 2} : - 2

\frac{- 5 - 3}{2 \cdot 2}

数字相减:

- 5 - 3 = - 8

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

二次方程组的解是:

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

u = cos (θ)

代回

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

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

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

cos (θ) = - 2 :

无解

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

无解

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

2 cos (x) = 2

sin(2θ)=sqrt(3)sin(θ)

sin (2 θ) = 3 sin (θ)

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

-8tan^2(θ)-4tan(θ)-11=6tan(θ)-8

- 8 tan^{2} (θ) - 4 tan (θ) - 11 = 6 tan (θ) - 8

2cos^2(x)+3cos(x)-5=0

2 cos^{2} (x) + 3 cos (x) - 5 = 0