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受欢迎的三角函数 >

-2cos(2θ)+9cos(θ)+4=cos(θ)+1

解答

- 2 cos (2 θ) + 9 cos (θ) + 4 = cos (θ) + 1

解答

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

+1

度数

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

求解步骤

- 2 cos (2 θ) + 9 cos (θ) + 4 = cos (θ) + 1

两边减去

cos (θ) + 1

8 cos (θ) - 2 cos (2 θ) + 3 = 0

使用三角恒等式改写

3 - 2 cos (2 θ) + 8 cos (θ)

使用倍角公式:

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

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

化简

3 - 2 (2 cos^{2} (θ) - 1) + 8 cos (θ) : 8 cos (θ) - 4 cos^{2} (θ) + 5

3 - 2 (2 cos^{2} (θ) - 1) + 8 cos (θ)

乘开

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

- 2 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 2 cos^{2} (θ), c = 1

= - 2 \cdot 2 cos^{2} (θ) - (- 2) \cdot 1

使用加减运算法则

- (- a) = a

= - 2 \cdot 2 cos^{2} (θ) + 2 \cdot 1

化简

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

- 2 \cdot 2 cos^{2} (θ) + 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= - 4 cos^{2} (θ) + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= - 4 cos^{2} (θ) + 2

= - 4 cos^{2} (θ) + 2

= 3 - 4 cos^{2} (θ) + 2 + 8 cos (θ)

化简

3 - 4 cos^{2} (θ) + 2 + 8 cos (θ) : 8 cos (θ) - 4 cos^{2} (θ) + 5

3 - 4 cos^{2} (θ) + 2 + 8 cos (θ)

对同类项分组

= - 4 cos^{2} (θ) + 8 cos (θ) + 3 + 2

数字相加:

3 + 2 = 5

= 8 cos (θ) - 4 cos^{2} (θ) + 5

= 8 cos (θ) - 4 cos^{2} (θ) + 5

= 8 cos (θ) - 4 cos^{2} (θ) + 5

5 - 4 cos^{2} (θ) + 8 cos (θ) = 0

用替代法求解

5 - 4 cos^{2} (θ) + 8 cos (θ) = 0

令

: cos (θ) = u

5 - 4 u^{2} + 8 u = 0

5 - 4 u^{2} + 8 u = 0 : u = - \frac{1}{2}, u = \frac{5}{2}

5 - 4 u^{2} + 8 u = 0

改写成标准形式

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

- 4 u^{2} + 8 u + 5 = 0

使用求根公式求解

- 4 u^{2} + 8 u + 5 = 0

二次方程求根公式:

若

a = - 4, b = 8, c = 5

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

8^{2} - 4 (- 4) \cdot 5 = 12

8^{2} - 4 (- 4) \cdot 5

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 4 \cdot 5

数字相乘:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

数字相加:

64 + 80 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- 8 \pm 12}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 8 + 12}{2 ( - 4 )}, u_{2} = \frac{- 8 - 12}{2 ( - 4 )}

u = \frac{- 8 + 12}{2 ( - 4 )} : - \frac{1}{2}

\frac{- 8 + 12}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 8 + 12}{- 2 \cdot 4}

数字相加/相减:

- 8 + 12 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = \frac{- 8 - 12}{2 ( - 4 )} : \frac{5}{2}

\frac{- 8 - 12}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 8 - 12}{- 2 \cdot 4}

数字相减:

- 8 - 12 = - 20

= \frac{- 20}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 20}{- 8}

使用分式法则:

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

= \frac{20}{8}

约分:

4

= \frac{5}{2}

二次方程组的解是:

u = - \frac{1}{2}, u = \frac{5}{2}

u = cos (θ)

代回

cos (θ) = - \frac{1}{2}, cos (θ) = \frac{5}{2}

cos (θ) = - \frac{1}{2}, cos (θ) = \frac{5}{2}

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{5}{2} :

无解

cos (θ) = \frac{5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

作图

流行的例子

csc^2(2x)-cot(2x)=1

csc^{2} (2 x) - cot (2 x) = 1

100tan(x)-8=50tan(x)

100 tan (x) - 8 = 50 tan (x)

cos(θ)+sin(θ)=1

cos (θ) + sin (θ) = 1

2 cos (3 x) = - 1

6sin(θ/2)=-6cos(θ/2)

6 sin (\frac{θ}{2}) = - 6 cos (\frac{θ}{2})