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受欢迎的三角函数 >

4cos(2θ)+1=2cos(θ)

解答

4 cos (2 θ) + 1 = 2 cos (θ)

解答

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

+1

度数

θ = 41.40962 \dots^{\circ} + 36 0^{\circ} n, θ = 318.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (2 θ) + 1 = 2 cos (θ)

两边减去

2 cos (θ)

4 cos (2 θ) + 1 - 2 cos (θ) = 0

使用三角恒等式改写

1 - 2 cos (θ) + 4 cos (2 θ)

使用倍角公式:

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

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

化简

1 - 2 cos (θ) + 4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 2 cos (θ) - 3

1 - 2 cos (θ) + 4 (2 cos^{2} (θ) - 1)

乘开

4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 4

4 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

化简

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1 : 8 cos^{2} (θ) - 4

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

数字相乘:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 1 - 2 cos (θ) + 8 cos^{2} (θ) - 4

化简

1 - 2 cos (θ) + 8 cos^{2} (θ) - 4 : 8 cos^{2} (θ) - 2 cos (θ) - 3

1 - 2 cos (θ) + 8 cos^{2} (θ) - 4

对同类项分组

= - 2 cos (θ) + 8 cos^{2} (θ) + 1 - 4

数字相加/相减:

1 - 4 = - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

= 8 cos^{2} (θ) - 2 cos (θ) - 3

- 3 - 2 cos (θ) + 8 cos^{2} (θ) = 0

用替代法求解

- 3 - 2 cos (θ) + 8 cos^{2} (θ) = 0

令

: cos (θ) = u

- 3 - 2 u + 8 u^{2} = 0

- 3 - 2 u + 8 u^{2} = 0 : u = \frac{3}{4}, u = - \frac{1}{2}

- 3 - 2 u + 8 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 8, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 3) = 10

(- 2)^{2} - 4 \cdot 8 (- 3)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 3

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

数字相加:

4 + 96 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 2 ) \pm 10}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 10}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 10}{2 \cdot 8}

u = \frac{- ( - 2 ) + 10}{2 \cdot 8} : \frac{3}{4}

\frac{- ( - 2 ) + 10}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{2 + 10}{2 \cdot 8}

数字相加:

2 + 10 = 12

= \frac{12}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{12}{16}

约分:

4

= \frac{3}{4}

u = \frac{- ( - 2 ) - 10}{2 \cdot 8} : - \frac{1}{2}

\frac{- ( - 2 ) - 10}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{2 - 10}{2 \cdot 8}

数字相减:

2 - 10 = - 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 8}{16}

使用分式法则:

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

= - \frac{8}{16}

约分:

8

= - \frac{1}{2}

二次方程组的解是:

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

u = cos (θ)

代回

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

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

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

作图

流行的例子

2sin^3(x)=2sin(x)

2 sin^{3} (x) = 2 sin (x)

sin^{4} (x) = 0

tan (θ) = \frac{7}{24}

-2sqrt(2)=-2sec(x+150)

- 22 = - 2 sec (x + 15 0^{\circ})

9sin(θ)+sqrt(11)=0

9 sin (θ) + 11 = 0