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受欢迎的三角函数 >

2cos(2θ)-7cos(θ)+2=-4cos(θ)

解答

2 cos (2 θ) - 7 cos (θ) + 2 = - 4 cos (θ)

解答

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 41.40962 \dots^{\circ} + 36 0^{\circ} n, θ = 318.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (2 θ) - 7 cos (θ) + 2 = - 4 cos (θ)

两边减去

- 4 cos (θ)

2 cos (2 θ) - 3 cos (θ) + 2 = 0

使用三角恒等式改写

2 + 2 cos (2 θ) - 3 cos (θ)

使用倍角公式:

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

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

化简

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

2 + 2 (2 cos^{2} (θ) - 1) - 3 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

= 2 + 4 cos^{2} (θ) - 2 - 3 cos (θ)

化简

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

2 + 4 cos^{2} (θ) - 2 - 3 cos (θ)

对同类项分组

= 4 cos^{2} (θ) - 3 cos (θ) + 2 - 2

2 - 2 = 0

= 4 cos^{2} (θ) - 3 cos (θ)

= 4 cos^{2} (θ) - 3 cos (θ)

= 4 cos^{2} (θ) - 3 cos (θ)

- 3 cos (θ) + 4 cos^{2} (θ) = 0

用替代法求解

- 3 cos (θ) + 4 cos^{2} (θ) = 0

令

: cos (θ) = u

- 3 u + 4 u^{2} = 0

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

- 3 u + 4 u^{2} = 0

改写成标准形式

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

4 u^{2} - 3 u = 0

使用求根公式求解

4 u^{2} - 3 u = 0

二次方程求根公式:

若

a = 4, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 4 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 4}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 4}

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

\frac{- ( - 3 ) + 3}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{3 + 3}{2 \cdot 4}

数字相加:

3 + 3 = 6

= \frac{6}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{6}{8}

约分:

2

= \frac{3}{4}

u = \frac{- ( - 3 ) - 3}{2 \cdot 4} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{3 - 3}{2 \cdot 4}

数字相减:

3 - 3 = 0

= \frac{0}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{0}{8}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = \frac{3}{4}, u = 0

u = cos (θ)

代回

cos (θ) = \frac{3}{4}, cos (θ) = 0

cos (θ) = \frac{3}{4}, cos (θ) = 0

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

cos (θ) = \frac{3}{4}

使用反三角函数性质

cos (θ) = \frac{3}{4}

cos (θ) = \frac{3}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 0

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

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

合并所有解

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

以小数形式表示解

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

sin(5x-15)=sin(5*x)

sin (5 x - 15) = sin (5 \cdot x)

3cos(x)-1=4cos(x)

3 cos (x) - 1 = 4 cos (x)

sin (x) = 666

-2cot(θ)=cot^2(θ)+1

- 2 cot (θ) = cot^{2} (θ) + 1

[sin(4x)cos(x)-cos(4x)sin(x)]^2=1

[sin (4 x) cos (x) - cos (4 x) sin (x)]^{2} = 1