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受欢迎的三角函数 >

-9sin^2(x)-5cos(x)+5=0

解答

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

解答

x = 2 πn, x = 2.03135 \dots + 2 πn, x = - 2.03135 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 116.38779 \dots^{\circ} + 36 0^{\circ} n, x = - 116.38779 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

5 - 5 cos (x) - 9 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)

= 5 - 5 cos (x) - 9 (1 - cos^{2} (x))

化简

5 - 5 cos (x) - 9 (1 - cos^{2} (x)) : 9 cos^{2} (x) - 5 cos (x) - 4

5 - 5 cos (x) - 9 (1 - cos^{2} (x))

乘开

- 9 (1 - cos^{2} (x)) : - 9 + 9 cos^{2} (x)

- 9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = cos^{2} (x)

= - 9 \cdot 1 - (- 9) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 cos^{2} (x)

= 5 - 5 cos (x) - 9 + 9 cos^{2} (x)

化简

5 - 5 cos (x) - 9 + 9 cos^{2} (x) : 9 cos^{2} (x) - 5 cos (x) - 4

5 - 5 cos (x) - 9 + 9 cos^{2} (x)

对同类项分组

= - 5 cos (x) + 9 cos^{2} (x) + 5 - 9

数字相加/相减:

5 - 9 = - 4

= 9 cos^{2} (x) - 5 cos (x) - 4

= 9 cos^{2} (x) - 5 cos (x) - 4

= 9 cos^{2} (x) - 5 cos (x) - 4

- 4 - 5 cos (x) + 9 cos^{2} (x) = 0

用替代法求解

- 4 - 5 cos (x) + 9 cos^{2} (x) = 0

令

: cos (x) = u

- 4 - 5 u + 9 u^{2} = 0

- 4 - 5 u + 9 u^{2} = 0 : u = 1, u = - \frac{4}{9}

- 4 - 5 u + 9 u^{2} = 0

改写成标准形式

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

9 u^{2} - 5 u - 4 = 0

使用求根公式求解

9 u^{2} - 5 u - 4 = 0

二次方程求根公式:

若

a = 9, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

(- 5)^{2} - 4 \cdot 9 (- 4) = 13

(- 5)^{2} - 4 \cdot 9 (- 4)

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 9 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 9 \cdot 4

数字相乘:

4 \cdot 9 \cdot 4 = 144

= 5^{2} + 144

5^{2} = 25

= 25 + 144

数字相加:

25 + 144 = 169

= 169

因式分解数字:

169 = 1 3^{2}

= 1 3^{2}

使用根式运算法则:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 9}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 9}, u_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 9}

u = \frac{- ( - 5 ) + 13}{2 \cdot 9} : 1

\frac{- ( - 5 ) + 13}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{5 + 13}{2 \cdot 9}

数字相加:

5 + 13 = 18

= \frac{18}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{18}{18}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 13}{2 \cdot 9} : - \frac{4}{9}

\frac{- ( - 5 ) - 13}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{5 - 13}{2 \cdot 9}

数字相减:

5 - 13 = - 8

= \frac{- 8}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{- 8}{18}

使用分式法则:

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

= - \frac{8}{18}

约分:

2

= - \frac{4}{9}

二次方程组的解是:

u = 1, u = - \frac{4}{9}

u = cos (x)

代回

cos (x) = 1, cos (x) = - \frac{4}{9}

cos (x) = 1, cos (x) = - \frac{4}{9}

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - \frac{4}{9} : x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

cos (x) = - \frac{4}{9}

使用反三角函数性质

cos (x) = - \frac{4}{9}

cos (x) = - \frac{4}{9}

的通解

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

x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

合并所有解

x = 2 πn, x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

以小数形式表示解

x = 2 πn, x = 2.03135 \dots + 2 πn, x = - 2.03135 \dots + 2 πn

作图

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