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受欢迎的三角函数 >

cos(2θ)=11sin(θ)-5

解答

cos (2 θ) = 11 sin (θ) - 5

解答

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

+1

度数

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

求解步骤

cos (2 θ) = 11 sin (θ) - 5

两边减去

11 sin (θ) - 5

cos (2 θ) - 11 sin (θ) + 5 = 0

使用三角恒等式改写

5 + cos (2 θ) - 11 sin (θ)

使用倍角公式:

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

= 5 + 1 - 2 sin^{2} (θ) - 11 sin (θ)

化简

= - 2 sin^{2} (θ) - 11 sin (θ) + 6

6 - 11 sin (θ) - 2 sin^{2} (θ) = 0

用替代法求解

6 - 11 sin (θ) - 2 sin^{2} (θ) = 0

令

: sin (θ) = u

6 - 11 u - 2 u^{2} = 0

6 - 11 u - 2 u^{2} = 0 : u = - 6, u = \frac{1}{2}

6 - 11 u - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} - 11 u + 6 = 0

使用求根公式求解

- 2 u^{2} - 11 u + 6 = 0

二次方程求根公式:

若

a = - 2, b = - 11, c = 6

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 2 ) \cdot 6}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 2 ) \cdot 6}{2 ( - 2 )}

(- 11)^{2} - 4 (- 2) \cdot 6 = 13

(- 11)^{2} - 4 (- 2) \cdot 6

使用法则

- (- a) = a

= (- 11)^{2} + 4 \cdot 2 \cdot 6

使用指数法则:

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

若

n

是偶数

(- 11)^{2} = 1 1^{2}

= 1 1^{2} + 4 \cdot 2 \cdot 6

数字相乘:

4 \cdot 2 \cdot 6 = 48

= 1 1^{2} + 48

1 1^{2} = 121

= 121 + 48

数字相加:

121 + 48 = 169

= 169

因式分解数字:

169 = 1 3^{2}

= 1 3^{2}

使用根式运算法则:

n a^{n} = a

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 11 ) \pm 13}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 11 ) + 13}{2 ( - 2 )}, u_{2} = \frac{- ( - 11 ) - 13}{2 ( - 2 )}

u = \frac{- ( - 11 ) + 13}{2 ( - 2 )} : - 6

\frac{- ( - 11 ) + 13}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{11 + 13}{- 2 \cdot 2}

数字相加:

11 + 13 = 24

= \frac{24}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{24}{- 4}

使用分式法则:

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

= - \frac{24}{4}

数字相除:

\frac{24}{4} = 6

= - 6

u = \frac{- ( - 11 ) - 13}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 11 ) - 13}{2 ( - 2 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{11 - 13}{- 2 \cdot 2}

数字相减:

11 - 13 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - 6, sin (θ) = \frac{1}{2}

sin (θ) = - 6, sin (θ) = \frac{1}{2}

sin (θ) = - 6 :

无解

sin (θ) = - 6

- 1 \leq sin (x) \leq 1

无解

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

作图

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