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受欢迎的三角函数 >

4cos(2θ)+3cos(θ)+4=-3cos(θ)-1

解答

4 cos (2 θ) + 3 cos (θ) + 4 = - 3 cos (θ) - 1

解答

θ = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

+1

度数

θ = 104.47751 \dots^{\circ} + 36 0^{\circ} n, θ = - 104.47751 \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) + 3 cos (θ) + 4 = - 3 cos (θ) - 1

两边减去

- 3 cos (θ) - 1

4 cos (2 θ) + 6 cos (θ) + 5 = 0

使用三角恒等式改写

5 + 4 cos (2 θ) + 6 cos (θ)

使用倍角公式:

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

= 5 + 4 (2 cos^{2} (θ) - 1) + 6 cos (θ)

化简

5 + 4 (2 cos^{2} (θ) - 1) + 6 cos (θ) : 8 cos^{2} (θ) + 6 cos (θ) + 1

5 + 4 (2 cos^{2} (θ) - 1) + 6 cos (θ)

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

= 5 + 8 cos^{2} (θ) - 4 + 6 cos (θ)

化简

5 + 8 cos^{2} (θ) - 4 + 6 cos (θ) : 8 cos^{2} (θ) + 6 cos (θ) + 1

5 + 8 cos^{2} (θ) - 4 + 6 cos (θ)

对同类项分组

= 8 cos^{2} (θ) + 6 cos (θ) + 5 - 4

数字相加/相减:

5 - 4 = 1

= 8 cos^{2} (θ) + 6 cos (θ) + 1

= 8 cos^{2} (θ) + 6 cos (θ) + 1

= 8 cos^{2} (θ) + 6 cos (θ) + 1

1 + 6 cos (θ) + 8 cos^{2} (θ) = 0

用替代法求解

1 + 6 cos (θ) + 8 cos^{2} (θ) = 0

令

: cos (θ) = u

1 + 6 u + 8 u^{2} = 0

1 + 6 u + 8 u^{2} = 0 : u = - \frac{1}{4}, u = - \frac{1}{2}

1 + 6 u + 8 u^{2} = 0

改写成标准形式

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

8 u^{2} + 6 u + 1 = 0

使用求根公式求解

8 u^{2} + 6 u + 1 = 0

二次方程求根公式:

若

a = 8, b = 6, c = 1

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

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

6^{2} - 4 \cdot 8 \cdot 1

数字相乘:

4 \cdot 8 \cdot 1 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

数字相减:

36 - 32 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

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

将解分隔开

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

u = \frac{- 6 + 2}{2 \cdot 8} : - \frac{1}{4}

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

数字相加/相减:

- 6 + 2 = - 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}

u = \frac{- 6 - 2}{2 \cdot 8} : - \frac{1}{2}

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

数字相减:

- 6 - 2 = - 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{1}{4}, u = - \frac{1}{2}

u = cos (θ)

代回

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

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

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

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

θ = arccos (- \frac{1}{4}) + 2 πn, θ = - arccos (- \frac{1}{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{1}{4}) + 2 πn, θ = - arccos (- \frac{1}{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 = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn, θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

作图

流行的例子

cos(θ-pi/4)-1=0

cos (θ - \frac{π}{4}) - 1 = 0

solvefor x,-sin(x)=0

so l v e f or x, - sin (x) = 0

2sec^2(x)tan(x)=0

2 sec^{2} (x) tan (x) = 0

2sin^2(x)-5sin(x)=-2

2 sin^{2} (x) - 5 sin (x) = - 2

8sin^2(x)+6sin(x)+1=0

8 sin^{2} (x) + 6 sin (x) + 1 = 0