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受欢迎的三角函数 >

3sin^2(x)=8cos(x)

解答

3 sin^{2} (x) = 8 cos (x)

解答

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度数

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin^{2} (x) = 8 cos (x)

两边减去

8 cos (x)

3 sin^{2} (x) - 8 cos (x) = 0

使用三角恒等式改写

3 sin^{2} (x) - 8 cos (x)

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} (x)) - 8 cos (x)

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

用替代法求解

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

令

: cos (x) = u

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

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

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 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 )} : - 3

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

去除括号:

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

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

数字相加:

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

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

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

去除括号:

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

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

数字相减:

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}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - 3 :

无解

cos (x) = - 3

- 1 \leq cos (x) \leq 1

无解

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

cos (x) = \frac{1}{3}

使用反三角函数性质

cos (x) = \frac{1}{3}

cos (x) = \frac{1}{3}

的通解

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

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

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

合并所有解

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

以小数形式表示解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

作图

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