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受欢迎的三角函数 >

-6cos^2(θ)=-cos(θ)-2

解答

- 6 cos^{2} (θ) = - cos (θ) - 2

解答

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

+1

度数

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 6 cos^{2} (θ) = - cos (θ) - 2

用替代法求解

- 6 cos^{2} (θ) = - cos (θ) - 2

令

: cos (θ) = u

- 6 u^{2} = - u - 2

- 6 u^{2} = - u - 2 : u = - \frac{1}{2}, u = \frac{2}{3}

- 6 u^{2} = - u - 2

将

2

para o lado esquerdo

- 6 u^{2} = - u - 2

两边加上

2

- 6 u^{2} + 2 = - u - 2 + 2

化简

- 6 u^{2} + 2 = - u

- 6 u^{2} + 2 = - u

将

u

para o lado esquerdo

- 6 u^{2} + 2 = - u

两边加上

u

- 6 u^{2} + 2 + u = - u + u

化简

- 6 u^{2} + 2 + u = 0

- 6 u^{2} + 2 + u = 0

改写成标准形式

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

- 6 u^{2} + u + 2 = 0

使用求根公式求解

- 6 u^{2} + u + 2 = 0

二次方程求根公式:

若

a = - 6, b = 1, c = 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 2}{2 ( - 6 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 2}{2 ( - 6 )}

1^{2} - 4 (- 6) \cdot 2 = 7

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 1 + 7}{2 ( - 6 )}, u_{2} = \frac{- 1 - 7}{2 ( - 6 )}

u = \frac{- 1 + 7}{2 ( - 6 )} : - \frac{1}{2}

\frac{- 1 + 7}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 + 7}{- 2 \cdot 6}

数字相加/相减:

- 1 + 7 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- 1 - 7}{2 ( - 6 )} : \frac{2}{3}

\frac{- 1 - 7}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 - 7}{- 2 \cdot 6}

数字相减:

- 1 - 7 = - 8

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

使用分式法则:

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

= \frac{8}{12}

约分:

4

= \frac{2}{3}

二次方程组的解是:

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

u = cos (θ)

代回

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

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

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

cos (θ) = \frac{2}{3}

使用反三角函数性质

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

的通解

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

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

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

合并所有解

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = arccos (\frac{2}{3}) + 2 πn, θ = 2 π - arccos (\frac{2}{3}) + 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 = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

作图

流行的例子

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

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tan(x)+cot(x)= 5/2

tan (x) + cot (x) = \frac{5}{2}

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

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

csc(2x)=-sqrt(2)

csc (2 x) = - 2

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