zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

10cos^2(x)+cos(x)=11sin^2(x)-9

解答

10 cos^{2} (x) + cos (x) = 11 sin^{2} (x) - 9

解答

x = 1.28104 \dots + 2 πn, x = 2 π - 1.28104 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

+1

度数

x = 73.39845 \dots^{\circ} + 36 0^{\circ} n, x = 286.60154 \dots^{\circ} + 36 0^{\circ} n, x = 109.47122 \dots^{\circ} + 36 0^{\circ} n, x = - 109.47122 \dots^{\circ} + 36 0^{\circ} n

求解步骤

10 cos^{2} (x) + cos (x) = 11 sin^{2} (x) - 9

两边减去

11 sin^{2} (x) - 9

10 cos^{2} (x) + cos (x) - 11 sin^{2} (x) + 9 = 0

使用三角恒等式改写

9 + cos (x) + 10 cos^{2} (x) - 11 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)

= 9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x))

化简

9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x)) : 21 cos^{2} (x) + cos (x) - 2

9 + cos (x) + 10 cos^{2} (x) - 11 (1 - cos^{2} (x))

乘开

- 11 (1 - cos^{2} (x)) : - 11 + 11 cos^{2} (x)

- 11 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 11, b = 1, c = cos^{2} (x)

= - 11 \cdot 1 - (- 11) cos^{2} (x)

使用加减运算法则

- (- a) = a

= - 11 \cdot 1 + 11 cos^{2} (x)

数字相乘:

11 \cdot 1 = 11

= - 11 + 11 cos^{2} (x)

= 9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x)

化简

9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x) : 21 cos^{2} (x) + cos (x) - 2

9 + cos (x) + 10 cos^{2} (x) - 11 + 11 cos^{2} (x)

对同类项分组

= cos (x) + 10 cos^{2} (x) + 11 cos^{2} (x) + 9 - 11

同类项相加:

10 cos^{2} (x) + 11 cos^{2} (x) = 21 cos^{2} (x)

= cos (x) + 21 cos^{2} (x) + 9 - 11

数字相加/相减:

9 - 11 = - 2

= 21 cos^{2} (x) + cos (x) - 2

= 21 cos^{2} (x) + cos (x) - 2

= 21 cos^{2} (x) + cos (x) - 2

- 2 + cos (x) + 21 cos^{2} (x) = 0

用替代法求解

- 2 + cos (x) + 21 cos^{2} (x) = 0

令

: cos (x) = u

- 2 + u + 21 u^{2} = 0

- 2 + u + 21 u^{2} = 0 : u = \frac{2}{7}, u = - \frac{1}{3}

- 2 + u + 21 u^{2} = 0

改写成标准形式

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

21 u^{2} + u - 2 = 0

使用求根公式求解

21 u^{2} + u - 2 = 0

二次方程求根公式:

若

a = 21, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 21 ( - 2 )}{2 \cdot 21}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 21 ( - 2 )}{2 \cdot 21}

1^{2} - 4 \cdot 21 (- 2) = 13

1^{2} - 4 \cdot 21 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 21 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 21 \cdot 2

数字相乘:

4 \cdot 21 \cdot 2 = 168

= 1 + 168

数字相加:

1 + 168 = 169

= 169

因式分解数字:

169 = 1 3^{2}

= 1 3^{2}

使用根式运算法则:

n a^{n} = a

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 21}

将解分隔开

u_{1} = \frac{- 1 + 13}{2 \cdot 21}, u_{2} = \frac{- 1 - 13}{2 \cdot 21}

u = \frac{- 1 + 13}{2 \cdot 21} : \frac{2}{7}

\frac{- 1 + 13}{2 \cdot 21}

数字相加/相减:

- 1 + 13 = 12

= \frac{12}{2 \cdot 21}

数字相乘:

2 \cdot 21 = 42

= \frac{12}{42}

约分:

6

= \frac{2}{7}

u = \frac{- 1 - 13}{2 \cdot 21} : - \frac{1}{3}

\frac{- 1 - 13}{2 \cdot 21}

数字相减:

- 1 - 13 = - 14

= \frac{- 14}{2 \cdot 21}

数字相乘:

2 \cdot 21 = 42

= \frac{- 14}{42}

使用分式法则:

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

= - \frac{14}{42}

约分:

14

= - \frac{1}{3}

二次方程组的解是:

u = \frac{2}{7}, u = - \frac{1}{3}

u = cos (x)

代回

cos (x) = \frac{2}{7}, cos (x) = - \frac{1}{3}

cos (x) = \frac{2}{7}, cos (x) = - \frac{1}{3}

cos (x) = \frac{2}{7} : x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

cos (x) = \frac{2}{7}

使用反三角函数性质

cos (x) = \frac{2}{7}

cos (x) = \frac{2}{7}

的通解

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

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn

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

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

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

合并所有解

x = arccos (\frac{2}{7}) + 2 πn, x = 2 π - arccos (\frac{2}{7}) + 2 πn, x = arccos (- \frac{1}{3}) + 2 πn, x = - arccos (- \frac{1}{3}) + 2 πn

以小数形式表示解

x = 1.28104 \dots + 2 πn, x = 2 π - 1.28104 \dots + 2 πn, x = 1.91063 \dots + 2 πn, x = - 1.91063 \dots + 2 πn

作图

流行的例子

sin(6.5x+2.5)=0.0851

sin (6.5 x + 2.5) = 0.0851

1-tan(x)=(-1)/3

1 - tan (x) = \frac{- 1}{3}

sin(x/3)=cos(x/2)

sin (\frac{x}{3}) = cos (\frac{x}{2})

cos(x)=(cot^2(x))/(csc^2(x))

cos (x) = \frac{cot ^{2} ( x )}{csc ^{2} ( x )}

2tan^2(x)-3cot^2(x)=5

2 tan^{2} (x) - 3 cot^{2} (x) = 5