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受欢迎的三角函数 >

-3cos(2θ)+2sin(θ)+5=9sin(θ)

解答

- 3 cos (2 θ) + 2 sin (θ) + 5 = 9 sin (θ)

解答

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 138.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

- 3 cos (2 θ) + 2 sin (θ) + 5 = 9 sin (θ)

两边减去

9 sin (θ)

- 3 cos (2 θ) - 7 sin (θ) + 5 = 0

使用三角恒等式改写

5 - 3 cos (2 θ) - 7 sin (θ)

使用倍角公式:

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

= 5 - 3 (1 - 2 sin^{2} (θ)) - 7 sin (θ)

化简

5 - 3 (1 - 2 sin^{2} (θ)) - 7 sin (θ) : 6 sin^{2} (θ) - 7 sin (θ) + 2

5 - 3 (1 - 2 sin^{2} (θ)) - 7 sin (θ)

乘开

- 3 (1 - 2 sin^{2} (θ)) : - 3 + 6 sin^{2} (θ)

- 3 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = 2 sin^{2} (θ)

= - 3 \cdot 1 - (- 3) \cdot 2 sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

化简

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ) : - 3 + 6 sin^{2} (θ)

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 2 = 6

= - 3 + 6 sin^{2} (θ)

= - 3 + 6 sin^{2} (θ)

= 5 - 3 + 6 sin^{2} (θ) - 7 sin (θ)

数字相减:

5 - 3 = 2

= 6 sin^{2} (θ) - 7 sin (θ) + 2

= 6 sin^{2} (θ) - 7 sin (θ) + 2

2 + 6 sin^{2} (θ) - 7 sin (θ) = 0

用替代法求解

2 + 6 sin^{2} (θ) - 7 sin (θ) = 0

令

: sin (θ) = u

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

2 + 6 u^{2} - 7 u = 0 : u = \frac{2}{3}, u = \frac{1}{2}

2 + 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 + 2 = 0

使用求根公式求解

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

二次方程求根公式:

若

a = 6, b = - 7, c = 2

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

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

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

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

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 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

7 + 1 = 8

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

数字相乘:

2 \cdot 6 = 12

= \frac{8}{12}

约分:

4

= \frac{2}{3}

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

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

使用法则

- (- a) = a

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

数字相减:

7 - 1 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

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

u = sin (θ)

代回

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

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

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

sin (θ) = \frac{2}{3}

使用反三角函数性质

sin (θ) = \frac{2}{3}

sin (θ) = \frac{2}{3}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

合并所有解

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

以小数形式表示解

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

流行的例子

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