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受欢迎的三角函数 >

-4cos(2θ)+7cos(θ)+4=9cos(θ)+7

解答

- 4 cos (2 θ) + 7 cos (θ) + 4 = 9 cos (θ) + 7

解答

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

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 4 cos (2 θ) + 7 cos (θ) + 4 = 9 cos (θ) + 7

两边减去

9 cos (θ) + 7

- 4 cos (2 θ) - 2 cos (θ) - 3 = 0

使用三角恒等式改写

- 3 - 2 cos (θ) - 4 cos (2 θ)

使用倍角公式:

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

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

化简

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

- 3 - 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

使用加减运算法则

- (- a) = a

= - 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

= - 3 - 2 cos (θ) - 8 cos^{2} (θ) + 4

化简

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

- 3 - 2 cos (θ) - 8 cos^{2} (θ) + 4

对同类项分组

= - 2 cos (θ) - 8 cos^{2} (θ) - 3 + 4

数字相加/相减:

- 3 + 4 = 1

= - 8 cos^{2} (θ) - 2 cos (θ) + 1

= - 8 cos^{2} (θ) - 2 cos (θ) + 1

= - 8 cos^{2} (θ) - 2 cos (θ) + 1

1 - 2 cos (θ) - 8 cos^{2} (θ) = 0

用替代法求解

1 - 2 cos (θ) - 8 cos^{2} (θ) = 0

令

: cos (θ) = u

1 - 2 u - 8 u^{2} = 0

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

1 - 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 + 1 = 0

使用求根公式求解

- 8 u^{2} - 2 u + 1 = 0

二次方程求根公式:

若

a = - 8, b = - 2, c = 1

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

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

(- 2)^{2} - 4 (- 8) \cdot 1 = 6

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

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 1

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 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数字相加:

4 + 32 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 6}{2 ( - 8 )}, u_{2} = \frac{- ( - 2 ) - 6}{2 ( - 8 )}

u = \frac{- ( - 2 ) + 6}{2 ( - 8 )} : - \frac{1}{2}

\frac{- ( - 2 ) + 6}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 + 6}{- 2 \cdot 8}

数字相加:

2 + 6 = 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}

u = \frac{- ( - 2 ) - 6}{2 ( - 8 )} : \frac{1}{4}

\frac{- ( - 2 ) - 6}{2 ( - 8 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{2 - 6}{- 2 \cdot 8}

数字相减:

2 - 6 = - 4

= \frac{- 4}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4}{- 16}

使用分式法则:

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

= \frac{4}{16}

约分:

4

= \frac{1}{4}

二次方程组的解是:

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

u = cos (θ)

代回

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

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

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) = \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

合并所有解

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

以小数形式表示解

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

作图

流行的例子

8cos^2(x)+2sin(x)-7=0

8 cos^{2} (x) + 2 sin (x) - 7 = 0

cos(θ)=sin(θ/3-10)

cos (θ) = sin (\frac{θ}{3} - 1 0^{\circ})

tan (x) = sec (x)

tan (θ) = \frac{8}{6}

2cos(x)tan(x)+tan(x)=1+2cos(x)

2 cos (x) tan (x) + tan (x) = 1 + 2 cos (x)