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受欢迎的三角函数 >

6sin^2(x)+5cos(x)=7

解答

6 sin^{2} (x) + 5 cos (x) = 7

解答

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度数

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

6 sin^{2} (x) + 5 cos (x) = 7

两边减去

7

6 sin^{2} (x) + 5 cos (x) - 7 = 0

使用三角恒等式改写

- 7 + 5 cos (x) + 6 sin^{2} (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)

= - 7 + 5 cos (x) + 6 (1 - cos^{2} (x))

化简

- 7 + 5 cos (x) + 6 (1 - cos^{2} (x)) : 5 cos (x) - 6 cos^{2} (x) - 1

- 7 + 5 cos (x) + 6 (1 - cos^{2} (x))

乘开

6 (1 - cos^{2} (x)) : 6 - 6 cos^{2} (x)

6 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 6, b = 1, c = cos^{2} (x)

= 6 \cdot 1 - 6 cos^{2} (x)

数字相乘:

6 \cdot 1 = 6

= 6 - 6 cos^{2} (x)

= - 7 + 5 cos (x) + 6 - 6 cos^{2} (x)

化简

- 7 + 5 cos (x) + 6 - 6 cos^{2} (x) : 5 cos (x) - 6 cos^{2} (x) - 1

- 7 + 5 cos (x) + 6 - 6 cos^{2} (x)

对同类项分组

= 5 cos (x) - 6 cos^{2} (x) - 7 + 6

数字相加/相减:

- 7 + 6 = - 1

= 5 cos (x) - 6 cos^{2} (x) - 1

= 5 cos (x) - 6 cos^{2} (x) - 1

= 5 cos (x) - 6 cos^{2} (x) - 1

- 1 + 5 cos (x) - 6 cos^{2} (x) = 0

用替代法求解

- 1 + 5 cos (x) - 6 cos^{2} (x) = 0

令

: cos (x) = u

- 1 + 5 u - 6 u^{2} = 0

- 1 + 5 u - 6 u^{2} = 0 : u = \frac{1}{3}, u = \frac{1}{2}

- 1 + 5 u - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} + 5 u - 1 = 0

使用求根公式求解

- 6 u^{2} + 5 u - 1 = 0

二次方程求根公式:

若

a = - 6, b = 5, c = - 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 6 ) ( - 1 )}{2 ( - 6 )}

5^{2} - 4 (- 6) (- 1) = 1

5^{2} - 4 (- 6) (- 1)

使用法则

- (- a) = a

= 5^{2} - 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 5 + 1}{2 ( - 6 )}, u_{2} = \frac{- 5 - 1}{2 ( - 6 )}

u = \frac{- 5 + 1}{2 ( - 6 )} : \frac{1}{3}

\frac{- 5 + 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 5 + 1}{- 2 \cdot 6}

数字相加/相减:

- 5 + 1 = - 4

= \frac{- 4}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 4}{- 12}

使用分式法则:

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

= \frac{4}{12}

约分:

4

= \frac{1}{3}

u = \frac{- 5 - 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 5 - 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 5 - 1}{- 2 \cdot 6}

数字相减:

- 5 - 1 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

的通解

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

合并所有解

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{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, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

作图

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