zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的 >

2cos^4(x)+3sin^2(x)-2=0

解答

2 cos^{4} (x) + 3 sin^{2} (x) - 2 = 0

解答

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{4} (x) + 3 sin^{2} (x) - 2 = 0

使用三角恒等式改写

- 2 + 2 cos^{4} (x) + 3 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)

= - 2 + 2 cos^{4} (x) + 3 (1 - cos^{2} (x))

化简

- 2 + 2 cos^{4} (x) + 3 (1 - cos^{2} (x)) : 2 cos^{4} (x) - 3 cos^{2} (x) + 1

- 2 + 2 cos^{4} (x) + 3 (1 - cos^{2} (x))

乘开

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

3 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

= 3 \cdot 1 - 3 cos^{2} (x)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (x)

= - 2 + 2 cos^{4} (x) + 3 - 3 cos^{2} (x)

化简

- 2 + 2 cos^{4} (x) + 3 - 3 cos^{2} (x) : 2 cos^{4} (x) - 3 cos^{2} (x) + 1

- 2 + 2 cos^{4} (x) + 3 - 3 cos^{2} (x)

对同类项分组

= 2 cos^{4} (x) - 3 cos^{2} (x) - 2 + 3

数字相加/相减:

- 2 + 3 = 1

= 2 cos^{4} (x) - 3 cos^{2} (x) + 1

= 2 cos^{4} (x) - 3 cos^{2} (x) + 1

= 2 cos^{4} (x) - 3 cos^{2} (x) + 1

1 + 2 cos^{4} (x) - 3 cos^{2} (x) = 0

用替代法求解

1 + 2 cos^{4} (x) - 3 cos^{2} (x) = 0

令

: cos (x) = u

1 + 2 u^{4} - 3 u^{2} = 0

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

1 + 2 u^{4} - 3 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 3 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - 3 v + 1 = 0

解

2 v^{2} - 3 v + 1 = 0 : v = 1, v = \frac{1}{2}

2 v^{2} - 3 v + 1 = 0

使用求根公式求解

2 v^{2} - 3 v + 1 = 0

二次方程求根公式:

若

a = 2, b = - 3, c = 1

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

v_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, v_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

v = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

\frac{- ( - 3 ) + 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

数字相加:

3 + 1 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

v = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

数字相减:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

v = 1, v = \frac{1}{2}

v = 1, v = \frac{1}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解

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

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

解为

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

u = cos (x)

代回

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

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

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) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

合并所有解

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

以小数形式表示解

x = 2 πn, x = π + 2 πn, x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

流行的例子

cos^2(x)+4sin^2(x/2)=1

cos^{2} (x) + 4 sin^{2} (\frac{x}{2}) = 1

4cos(x)=sin^2(x)+1

4 cos (x) = sin^{2} (x) + 1

cos^2(x)+sin^2(x)-2sin(x)=0,0<= x<= 2pi

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

cot (A) = \frac{24}{7}

2sqrt(3)sec(θ)+4=0

23 sec (θ) + 4 = 0