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 >

sin^4(x)+sin^2(x)=sin^6(x)

解答

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

用替代法求解

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

令

: sin (x) = u

u^{4} + u^{2} = u^{6}

u^{4} + u^{2} = u^{6} : u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{4} + u^{2} = u^{6}

交换两边

u^{6} = u^{4} + u^{2}

将

u^{2}

para o lado esquerdo

u^{6} = u^{4} + u^{2}

两边减去

u^{2}

u^{6} - u^{2} = u^{4} + u^{2} - u^{2}

化简

u^{6} - u^{2} = u^{4}

u^{6} - u^{2} = u^{4}

将

u^{4}

para o lado esquerdo

u^{6} - u^{2} = u^{4}

两边减去

u^{4}

u^{6} - u^{2} - u^{4} = u^{4} - u^{4}

化简

u^{6} - u^{2} - u^{4} = 0

u^{6} - u^{2} - u^{4} = 0

改写成标准形式

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

u^{6} - u^{4} - u^{2} = 0

用

v = u^{2}, v^{2} = u^{4}

和

v^{3} = u^{6}

改写方程式

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

解

v^{3} - v^{2} - v = 0 : v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

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

因式分解

v^{3} - v^{2} - v : v (v^{2} - v - 1)

v^{3} - v^{2} - v

使用指数法则:

a^{b + c} = a^{b} a^{c}

v^{2} = vv

= v^{2} v - vv - v

因式分解出通项

v

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

v (v^{2} - v - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

v = 0 or v^{2} - v - 1 = 0

解

v^{2} - v - 1 = 0 : v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{2} - v - 1 = 0

使用求根公式求解

v^{2} - v - 1 = 0

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

将解分隔开

v_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

v = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

v = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

解为

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

解

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

u^{2} = \frac{1 + 5}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

解

u^{2} = \frac{1 - 5}{2} : u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{2} = \frac{1 - 5}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

解为

u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{1 - 5}{2}, sin (x) = - \frac{1 - 5}{2}

sin (x) = 0, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{1 - 5}{2}, sin (x) = - \frac{1 - 5}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

sin (x) = \frac{1 + 5}{2} :

无解

sin (x) = \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (x) = - \frac{1 + 5}{2} :

无解

sin (x) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{1 - 5}{2} : x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = \frac{1 - 5}{2}

使用反三角函数性质

sin (x) = \frac{1 - 5}{2}

sin (x) = \frac{1 - 5}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2} : x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2}

使用反三角函数性质

sin (x) = - \frac{1 - 5}{2}

sin (x) = - \frac{1 - 5}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn, x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

因为方程对以下值无定义:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

x = 2 πn, x = π + 2 πn

作图

流行的例子

cos(a)=(-11)/(14)

cos (a) = \frac{- 11}{14}

solvefor x,sin(x/x)=0.7 求解

x, sin (\frac{x}{x}) = 0.7

5cos^2(x)+sin^2(x)=4

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

cos(u)-1.5sin^2(u)+0.1667=0

cos (u) - 1.5 sin^{2} (u) + 0.1667 = 0

3 sin (2 x - 1) = 1