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受欢迎的三角函数 >

3sin^2(2θ)+8cos(2θ)=0

解答

3 sin^{2} (2 θ) + 8 cos (2 θ) = 0

解答

θ = \frac{1.91063 \dots}{2} + πn, θ = - \frac{1.91063 \dots}{2} + πn

+1

度数

θ = 54.73561 \dots^{\circ} + 18 0^{\circ} n, θ = - 54.73561 \dots^{\circ} + 18 0^{\circ} n

求解步骤

3 sin^{2} (2 θ) + 8 cos (2 θ) = 0

使用三角恒等式改写

3 sin^{2} (2 θ) + 8 cos (2 θ)

使用毕达哥拉斯恒等式:

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

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

= 3 (1 - cos^{2} (2 θ)) + 8 cos (2 θ)

(1 - cos^{2} (2 θ)) \cdot 3 + 8 cos (2 θ) = 0

用替代法求解

(1 - cos^{2} (2 θ)) \cdot 3 + 8 cos (2 θ) = 0

令

: cos (2 θ) = u

(1 - u^{2}) \cdot 3 + 8 u = 0

(1 - u^{2}) \cdot 3 + 8 u = 0 : u = - \frac{1}{3}, u = 3

(1 - u^{2}) \cdot 3 + 8 u = 0

展开

(1 - u^{2}) \cdot 3 + 8 u : 3 - 3 u^{2} + 8 u

(1 - u^{2}) \cdot 3 + 8 u

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

乘开

3 (1 - u^{2}) : 3 - 3 u^{2}

3 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

= 3 \cdot 1 - 3 u^{2}

数字相乘:

3 \cdot 1 = 3

= 3 - 3 u^{2}

= 3 - 3 u^{2} + 8 u

3 - 3 u^{2} + 8 u = 0

改写成标准形式

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

- 3 u^{2} + 8 u + 3 = 0

使用求根公式求解

- 3 u^{2} + 8 u + 3 = 0

二次方程求根公式:

若

a = - 3, b = 8, c = 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) \cdot 3}{2 ( - 3 )}

8^{2} - 4 (- 3) \cdot 3 = 10

8^{2} - 4 (- 3) \cdot 3

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 3 \cdot 3

数字相乘:

4 \cdot 3 \cdot 3 = 36

= 8^{2} + 36

8^{2} = 64

= 64 + 36

数字相加:

64 + 36 = 100

= 100

因式分解数字:

100 = 1 0^{2}

= 1 0^{2}

使用根式运算法则:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 8 \pm 10}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 8 + 10}{2 ( - 3 )}, u_{2} = \frac{- 8 - 10}{2 ( - 3 )}

u = \frac{- 8 + 10}{2 ( - 3 )} : - \frac{1}{3}

\frac{- 8 + 10}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 8 + 10}{- 2 \cdot 3}

数字相加/相减:

- 8 + 10 = 2

= \frac{2}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{2}{- 6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

u = \frac{- 8 - 10}{2 ( - 3 )} : 3

\frac{- 8 - 10}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 8 - 10}{- 2 \cdot 3}

数字相减:

- 8 - 10 = - 18

= \frac{- 18}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 18}{- 6}

使用分式法则:

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

= \frac{18}{6}

数字相除:

\frac{18}{6} = 3

= 3

二次方程组的解是:

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

u = cos (2 θ)

代回

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

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

cos (2 θ) = - \frac{1}{3} : θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn, θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

cos (2 θ) = - \frac{1}{3}

使用反三角函数性质

cos (2 θ) = - \frac{1}{3}

cos (2 θ) = - \frac{1}{3}

的通解

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

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

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

解

2 θ = arccos (- \frac{1}{3}) + 2 πn : θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn

2 θ = arccos (- \frac{1}{3}) + 2 πn

两边除以

2

2 θ = arccos (- \frac{1}{3}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( - \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn

θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn

解

2 θ = - arccos (- \frac{1}{3}) + 2 πn : θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

2 θ = - arccos (- \frac{1}{3}) + 2 πn

两边除以

2

2 θ = - arccos (- \frac{1}{3}) + 2 πn

两边除以

2

\frac{2 θ}{2} = - \frac{arccos ( - \frac{1}{3} )}{2} + \frac{2 πn}{2}

化简

θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

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

cos (2 θ) = 3 :

无解

cos (2 θ) = 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

以小数形式表示解

θ = \frac{1.91063 \dots}{2} + πn, θ = - \frac{1.91063 \dots}{2} + πn

作图

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