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受欢迎的三角函数 >

2cos(2θ)+2cos(θ)+2=3cos(θ)

解答

2 cos (2 θ) + 2 cos (θ) + 2 = 3 cos (θ)

解答

θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \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) + 2 cos (θ) + 2 = 3 cos (θ)

两边减去

3 cos (θ)

2 cos (2 θ) - cos (θ) + 2 = 0

使用三角恒等式改写

2 - cos (θ) + 2 cos (2 θ)

使用倍角公式:

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

= 2 - cos (θ) + 2 (2 cos^{2} (θ) - 1)

化简

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

2 - cos (θ) + 2 (2 cos^{2} (θ) - 1)

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 - cos (θ) + 4 cos^{2} (θ) - 2

化简

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

2 - cos (θ) + 4 cos^{2} (θ) - 2

对同类项分组

= - cos (θ) + 4 cos^{2} (θ) + 2 - 2

2 - 2 = 0

= 4 cos^{2} (θ) - cos (θ)

= 4 cos^{2} (θ) - cos (θ)

= 4 cos^{2} (θ) - cos (θ)

- cos (θ) + 4 cos^{2} (θ) = 0

用替代法求解

- cos (θ) + 4 cos^{2} (θ) = 0

令

: cos (θ) = u

- u + 4 u^{2} = 0

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

- u + 4 u^{2} = 0

改写成标准形式

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

4 u^{2} - u = 0

使用求根公式求解

4 u^{2} - u = 0

二次方程求根公式:

若

a = 4, b = - 1, c = 0

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

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

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

(- 1)^{2} - 4 \cdot 4 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

\frac{- ( - 1 ) + 1}{2 \cdot 4}

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{2}{8}

约分:

2

= \frac{1}{4}

u = \frac{- ( - 1 ) - 1}{2 \cdot 4} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 4}

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 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{1}{4}, u = 0

u = cos (θ)

代回

cos (θ) = \frac{1}{4}, cos (θ) = 0

cos (θ) = \frac{1}{4}, cos (θ) = 0

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

cos (θ) = \frac{1}{4}

使用反三角函数性质

cos (θ) = \frac{1}{4}

cos (θ) = \frac{1}{4}

的通解

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

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

θ = arccos (\frac{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{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{1}{4}) + 2 πn, θ = 2 π - arccos (\frac{1}{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 = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

流行的例子

1+csc(x)=cot^2(x)

1 + csc (x) = cot^{2} (x)

cos(1/(4θ))=(-sqrt(2))/2

cos (\frac{1}{4 θ}) = \frac{- 2}{2}

cos(x)tan(x)-2cos^2(x)=-1

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

sqrt(3)cot(1/4 x)=1

3 cot (\frac{1}{4} x) = 1

2sin(pi/4-x)+sqrt(3)=0

2 sin (\frac{π}{4} - x) + 3 = 0