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 >

csc^4(x)-4csc^2(x)=0

解答

csc^{4} (x) - 4 csc^{2} (x) = 0

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

csc^{4} (x) - 4 csc^{2} (x) = 0

用替代法求解

csc^{4} (x) - 4 csc^{2} (x) = 0

令

: csc (x) = u

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

u^{4} - 4 u^{2} = 0 : u = 2, u = - 2, u = 0

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 4 v = 0

解

v^{2} - 4 v = 0 : v = 4, v = 0

v^{2} - 4 v = 0

使用求根公式求解

v^{2} - 4 v = 0

二次方程求根公式:

若

a = 1, b = - 4, c = 0

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 0 = 4

(- 4)^{2} - 4 \cdot 1 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 4

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

将解分隔开

v_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 1}, v_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 1}

v = \frac{- ( - 4 ) + 4}{2 \cdot 1} : 4

\frac{- ( - 4 ) + 4}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 4}{2 \cdot 1}

数字相加:

4 + 4 = 8

= \frac{8}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

v = \frac{- ( - 4 ) - 4}{2 \cdot 1} : 0

\frac{- ( - 4 ) - 4}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 - 4}{2 \cdot 1}

数字相减:

4 - 4 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

v = 4, v = 0

v = 4, v = 0

代回

v = u^{2}

，求解

u

解

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

对于

x^{2} = f (a)

解为

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

u = 4, u = - 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

解为

u = 2, u = - 2, u = 0

u = csc (x)

代回

csc (x) = 2, csc (x) = - 2, csc (x) = 0

csc (x) = 2, csc (x) = - 2, csc (x) = 0

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

csc (x) = 2

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = - 2 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

csc (x) = - 2

csc (x) = - 2

的通解

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

csc (x) = 0 :

无解

csc (x) = 0

csc (x) \leq - 1 or csc (x) \geq 1

无解

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

作图

流行的例子

tan^2(x)-2sec(x)=-1

tan^{2} (x) - 2 sec (x) = - 1

cos(2θ)+14sin^2(θ)=10

cos (2 θ) + 14 sin^{2} (θ) = 10

cos^2(x)-1/4 =0

cos^{2} (x) - \frac{1}{4} = 0

3(1-sin(x))=2cos^2(x)

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

∣ cos (x) ∣ = \frac{1}{2}