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受欢迎的三角函数 >

4cos^2(θ)+cos(2θ)-7cos(θ)=-2

解答

4 cos^{2} (θ) + cos (2 θ) - 7 cos (θ) = - 2

解答

θ = 2 πn, θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 80.40593 \dots^{\circ} + 36 0^{\circ} n, θ = 279.59406 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos^{2} (θ) + cos (2 θ) - 7 cos (θ) = - 2

两边减去

- 2

4 cos^{2} (θ) + cos (2 θ) - 7 cos (θ) + 2 = 0

使用三角恒等式改写

2 + cos (2 θ) + 4 cos^{2} (θ) - 7 cos (θ)

使用倍角公式:

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

= 2 + 2 cos^{2} (θ) - 1 + 4 cos^{2} (θ) - 7 cos (θ)

化简

2 + 2 cos^{2} (θ) - 1 + 4 cos^{2} (θ) - 7 cos (θ) : 6 cos^{2} (θ) - 7 cos (θ) + 1

2 + 2 cos^{2} (θ) - 1 + 4 cos^{2} (θ) - 7 cos (θ)

对同类项分组

= 2 cos^{2} (θ) + 4 cos^{2} (θ) - 7 cos (θ) + 2 - 1

同类项相加:

2 cos^{2} (θ) + 4 cos^{2} (θ) = 6 cos^{2} (θ)

= 6 cos^{2} (θ) - 7 cos (θ) + 2 - 1

数字相加/相减:

2 - 1 = 1

= 6 cos^{2} (θ) - 7 cos (θ) + 1

= 6 cos^{2} (θ) - 7 cos (θ) + 1

1 + 6 cos^{2} (θ) - 7 cos (θ) = 0

用替代法求解

1 + 6 cos^{2} (θ) - 7 cos (θ) = 0

令

: cos (θ) = u

1 + 6 u^{2} - 7 u = 0

1 + 6 u^{2} - 7 u = 0 : u = 1, u = \frac{1}{6}

1 + 6 u^{2} - 7 u = 0

改写成标准形式

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

6 u^{2} - 7 u + 1 = 0

使用求根公式求解

6 u^{2} - 7 u + 1 = 0

二次方程求根公式:

若

a = 6, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 \cdot 1 = 5

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

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

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

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 6}

u = \frac{- ( - 7 ) + 5}{2 \cdot 6} : 1

\frac{- ( - 7 ) + 5}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 + 5}{2 \cdot 6}

数字相加:

7 + 5 = 12

= \frac{12}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{12}{12}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 7 ) - 5}{2 \cdot 6} : \frac{1}{6}

\frac{- ( - 7 ) - 5}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{7 - 5}{2 \cdot 6}

数字相减:

7 - 5 = 2

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

数字相乘:

2 \cdot 6 = 12

= \frac{2}{12}

约分:

2

= \frac{1}{6}

二次方程组的解是:

u = 1, u = \frac{1}{6}

u = cos (θ)

代回

cos (θ) = 1, cos (θ) = \frac{1}{6}

cos (θ) = 1, cos (θ) = \frac{1}{6}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = \frac{1}{6} : θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

cos (θ) = \frac{1}{6}

使用反三角函数性质

cos (θ) = \frac{1}{6}

cos (θ) = \frac{1}{6}

的通解

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

θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

合并所有解

θ = 2 πn, θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

以小数形式表示解

θ = 2 πn, θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn

作图

流行的例子

sec^2(θ)+2sec(θ)=8

sec^{2} (θ) + 2 sec (θ) = 8

8 tan (θ) + 9 = 0

1+tanh^2(x)=sech^2(x)

1 + tanh^{2} (x) = sech^{2} (x)

cos(x)-sin(x)= 1/2

cos (x) - sin (x) = \frac{1}{2}

cos (x) = - \frac{4}{9}