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受欢迎的三角函数 >

12sin^2(x)+7cos(x)-13=0

解答

12 sin^{2} (x) + 7 cos (x) - 13 = 0

解答

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度数

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n, x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

12 sin^{2} (x) + 7 cos (x) - 13 = 0

使用三角恒等式改写

- 13 + 12 sin^{2} (x) + 7 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)

= - 13 + 12 (1 - cos^{2} (x)) + 7 cos (x)

化简

- 13 + 12 (1 - cos^{2} (x)) + 7 cos (x) : 7 cos (x) - 12 cos^{2} (x) - 1

- 13 + 12 (1 - cos^{2} (x)) + 7 cos (x)

乘开

12 (1 - cos^{2} (x)) : 12 - 12 cos^{2} (x)

12 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 12, b = 1, c = cos^{2} (x)

= 12 \cdot 1 - 12 cos^{2} (x)

数字相乘:

12 \cdot 1 = 12

= 12 - 12 cos^{2} (x)

= - 13 + 12 - 12 cos^{2} (x) + 7 cos (x)

数字相加/相减:

- 13 + 12 = - 1

= 7 cos (x) - 12 cos^{2} (x) - 1

= 7 cos (x) - 12 cos^{2} (x) - 1

- 1 - 12 cos^{2} (x) + 7 cos (x) = 0

用替代法求解

- 1 - 12 cos^{2} (x) + 7 cos (x) = 0

令

: cos (x) = u

- 1 - 12 u^{2} + 7 u = 0

- 1 - 12 u^{2} + 7 u = 0 : u = \frac{1}{4}, u = \frac{1}{3}

- 1 - 12 u^{2} + 7 u = 0

改写成标准形式

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

- 12 u^{2} + 7 u - 1 = 0

使用求根公式求解

- 12 u^{2} + 7 u - 1 = 0

二次方程求根公式:

若

a = - 12, b = 7, c = - 1

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 12 ) ( - 1 )}{2 ( - 12 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 12 ) ( - 1 )}{2 ( - 12 )}

7^{2} - 4 (- 12) (- 1) = 1

7^{2} - 4 (- 12) (- 1)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 12 \cdot 1

数字相乘:

4 \cdot 12 \cdot 1 = 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 ( - 12 )}

将解分隔开

u_{1} = \frac{- 7 + 1}{2 ( - 12 )}, u_{2} = \frac{- 7 - 1}{2 ( - 12 )}

u = \frac{- 7 + 1}{2 ( - 12 )} : \frac{1}{4}

\frac{- 7 + 1}{2 ( - 12 )}

去除括号:

(- a) = - a

= \frac{- 7 + 1}{- 2 \cdot 12}

数字相加/相减:

- 7 + 1 = - 6

= \frac{- 6}{- 2 \cdot 12}

数字相乘:

2 \cdot 12 = 24

= \frac{- 6}{- 24}

使用分式法则:

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

= \frac{6}{24}

约分:

6

= \frac{1}{4}

u = \frac{- 7 - 1}{2 ( - 12 )} : \frac{1}{3}

\frac{- 7 - 1}{2 ( - 12 )}

去除括号:

(- a) = - a

= \frac{- 7 - 1}{- 2 \cdot 12}

数字相减:

- 7 - 1 = - 8

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

数字相乘:

2 \cdot 12 = 24

= \frac{- 8}{- 24}

使用分式法则:

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

= \frac{8}{24}

约分:

8

= \frac{1}{3}

二次方程组的解是:

u = \frac{1}{4}, u = \frac{1}{3}

u = cos (x)

代回

cos (x) = \frac{1}{4}, cos (x) = \frac{1}{3}

cos (x) = \frac{1}{4}, cos (x) = \frac{1}{3}

cos (x) = \frac{1}{4} : x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = \frac{1}{4}

使用反三角函数性质

cos (x) = \frac{1}{4}

cos (x) = \frac{1}{4}

的通解

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

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

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}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn, 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 = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

作图

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