zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的 >

6sin^2(x)=5+cos(x),180<= x<= 540

解答

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

解答

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - 1.23095 \dots, x = 1.23095 \dots + 36 0^{\circ}

+1

弧度

x = \frac{4 π}{3}, x = \frac{8 π}{3}, x = 2 π - 1.23095 \dots, x = 1.23095 \dots + 2 π

求解步骤

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

两边减去

5 + cos (x)

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

使用三角恒等式改写

- 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)

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

化简

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

- 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)

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 5 + 6 = 1

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

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

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

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

用替代法求解

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

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

(- 1)^{2} - 4 (- 6) \cdot 1

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 6 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 6 \cdot 1 = 24

4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

5^{2} = 5

= 5

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

1 + 5 = 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}

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

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

去除括号:

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

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

数字相减:

1 - 5 = - 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}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = - \frac{1}{2}, 18 0^{\circ} \leq x \leq 54 0^{\circ} : x = 24 0^{\circ}, x = 48 0^{\circ}

cos (x) = - \frac{1}{2}, 18 0^{\circ} \leq x \leq 54 0^{\circ}

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

的通解

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

在

18 0^{\circ} \leq x \leq 54 0^{\circ}

范围内的解

x = 24 0^{\circ}, x = 48 0^{\circ}

cos (x) = \frac{1}{3}, 18 0^{\circ} \leq x \leq 54 0^{\circ} : x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

cos (x) = \frac{1}{3}, 18 0^{\circ} \leq x \leq 54 0^{\circ}

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) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

x = arccos (\frac{1}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{3}) + 36 0^{\circ} n

在

18 0^{\circ} \leq x \leq 54 0^{\circ}

范围内的解

x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

合并所有解

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - arccos (\frac{1}{3}), x = arccos (\frac{1}{3}) + 36 0^{\circ}

以小数形式表示解

x = 24 0^{\circ}, x = 48 0^{\circ}, x = 36 0^{\circ} - 1.23095 \dots, x = 1.23095 \dots + 36 0^{\circ}

流行的例子

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

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

2sin(x)cos(2x)-cos(2x)=0

2 sin (x) cos (2 x) - cos (2 x) = 0

- 2 = cos (x)

2 cot (2 x) + 2 = 0

sin(x-pi)=(-sqrt(2))/2

sin (x - π) = \frac{- 2}{2}