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受欢迎的三角函数 >

2cos(2θ)+4cos(θ)+2=-1

解答

2 cos (2 θ) + 4 cos (θ) + 2 = - 1

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

2 cos (2 θ) + 4 cos (θ) + 2 = - 1

两边减去

- 1

2 cos (2 θ) + 4 cos (θ) + 3 = 0

使用三角恒等式改写

3 + 2 cos (2 θ) + 4 cos (θ)

使用倍角公式:

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

= 3 + 2 (2 cos^{2} (θ) - 1) + 4 cos (θ)

化简

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

3 + 2 (2 cos^{2} (θ) - 1) + 4 cos (θ)

乘开

2 (2 cos^{2} (θ) - 1) : 4 cos^{2} (θ) - 2

2 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (θ), c = 1

= 2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

化简

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

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 4 cos^{2} (θ) - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 4 cos^{2} (θ) - 2

= 4 cos^{2} (θ) - 2

= 3 + 4 cos^{2} (θ) - 2 + 4 cos (θ)

化简

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

3 + 4 cos^{2} (θ) - 2 + 4 cos (θ)

对同类项分组

= 4 cos^{2} (θ) + 4 cos (θ) + 3 - 2

数字相加/相减:

3 - 2 = 1

= 4 cos^{2} (θ) + 4 cos (θ) + 1

= 4 cos^{2} (θ) + 4 cos (θ) + 1

= 4 cos^{2} (θ) + 4 cos (θ) + 1

1 + 4 cos (θ) + 4 cos^{2} (θ) = 0

用替代法求解

1 + 4 cos (θ) + 4 cos^{2} (θ) = 0

令

: cos (θ) = u

1 + 4 u + 4 u^{2} = 0

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

1 + 4 u + 4 u^{2} = 0

改写成标准形式

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

4 u^{2} + 4 u + 1 = 0

使用求根公式求解

4 u^{2} + 4 u + 1 = 0

二次方程求根公式:

若

a = 4, b = 4, c = 1

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

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

4^{2} - 4 \cdot 4 \cdot 1 = 0

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 4}

u = \frac{- 4}{2 \cdot 4}

\frac{- 4}{2 \cdot 4} = - \frac{1}{2}

\frac{- 4}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = - \frac{1}{2}

二次方程组的解是:

u = - \frac{1}{2}

u = cos (θ)

代回

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{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

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

2cos(2x+(3pi)/2)=sqrt(3)

2 cos (2 x + \frac{3 π}{2}) = 3

3sin(x)+2cos(x)=0

3 sin (x) + 2 cos (x) = 0

2 = sec^{2} (x)

cosh (y) = 2

(sin(12))/(sin(x))= 343/1484

\frac{sin ( 1 2 ^{\circ} )}{sin ( x )} = \frac{343}{1484}