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 >

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

解答

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

解答

x = 1.13640 \dots + 2 πn, x = 2 π - 1.13640 \dots + 2 πn, x = 2.00519 \dots + 2 πn, x = - 2.00519 \dots + 2 πn

+1

度数

x = 65.11101 \dots^{\circ} + 36 0^{\circ} n, x = 294.88898 \dots^{\circ} + 36 0^{\circ} n, x = 114.88898 \dots^{\circ} + 36 0^{\circ} n, x = - 114.88898 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

2 u^{4} - 2 u^{2} = 4 u^{2} - 1

2 u^{4} - 2 u^{2} = 4 u^{2} - 1 : u = \frac{3 + 7}{2}, u = - \frac{3 + 7}{2}, u = \frac{3 - 7}{2}, u = - \frac{3 - 7}{2}

2 u^{4} - 2 u^{2} = 4 u^{2} - 1

将

1

para o lado esquerdo

2 u^{4} - 2 u^{2} = 4 u^{2} - 1

两边加上

1

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

化简

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

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

将

4 u^{2}

para o lado esquerdo

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

两边减去

4 u^{2}

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

化简

2 u^{4} - 6 u^{2} + 1 = 0

2 u^{4} - 6 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

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

解

2 v^{2} - 6 v + 1 = 0 : v = \frac{3 + 7}{2}, v = \frac{3 - 7}{2}

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = - 6, c = 1

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

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

(- 6)^{2} - 4 \cdot 2 \cdot 1 = 27

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

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 6^{2} - 8

6^{2} = 36

= 36 - 8

数字相减:

36 - 8 = 28

= 28

28

质因数分解:

2^{2} \cdot 7

28

28

除以

228 = 14 \cdot 2

= 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 27

v_{1, 2} = \frac{- ( - 6 ) \pm 2 7}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - 6 ) + 2 7}{2 \cdot 2}, v_{2} = \frac{- ( - 6 ) - 2 7}{2 \cdot 2}

v = \frac{- ( - 6 ) + 2 7}{2 \cdot 2} : \frac{3 + 7}{2}

\frac{- ( - 6 ) + 2 7}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{6 + 2 7}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{6 + 2 7}{4}

分解

6 + 27 : 2 (3 + 7)

6 + 27

改写为

= 2 \cdot 3 + 27

因式分解出通项

2

= 2 (3 + 7)

= \frac{2 ( 3 + 7 )}{4}

约分:

2

= \frac{3 + 7}{2}

v = \frac{- ( - 6 ) - 2 7}{2 \cdot 2} : \frac{3 - 7}{2}

\frac{- ( - 6 ) - 2 7}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{6 - 2 7}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{6 - 2 7}{4}

分解

6 - 27 : 2 (3 - 7)

6 - 27

改写为

= 2 \cdot 3 - 27

因式分解出通项

2

= 2 (3 - 7)

= \frac{2 ( 3 - 7 )}{4}

约分:

2

= \frac{3 - 7}{2}

二次方程组的解是:

v = \frac{3 + 7}{2}, v = \frac{3 - 7}{2}

v = \frac{3 + 7}{2}, v = \frac{3 - 7}{2}

代回

v = u^{2}

，求解

u

解

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

u^{2} = \frac{3 + 7}{2}

对于

x^{2} = f (a)

解为

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

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

解

u^{2} = \frac{3 - 7}{2} : u = \frac{3 - 7}{2}, u = - \frac{3 - 7}{2}

u^{2} = \frac{3 - 7}{2}

对于

x^{2} = f (a)

解为

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

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

解为

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

u = cos (x)

代回

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

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

cos (x) = \frac{3 + 7}{2} :

无解

cos (x) = \frac{3 + 7}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = - \frac{3 + 7}{2} :

无解

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

- 1 \leq cos (x) \leq 1

无解

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

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

使用反三角函数性质

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

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

的通解

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

x = arccos - \frac{3 - 7}{2} + 2 πn, x = - arccos - \frac{3 - 7}{2} + 2 πn

x = arccos - \frac{3 - 7}{2} + 2 πn, x = - arccos - \frac{3 - 7}{2} + 2 πn

合并所有解

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

以小数形式表示解

x = 1.13640 \dots + 2 πn, x = 2 π - 1.13640 \dots + 2 πn, x = 2.00519 \dots + 2 πn, x = - 2.00519 \dots + 2 πn

作图

流行的例子

solvefor x,t^2=2sec(x)

so l v e f or x, t^{2} = 2 sec (x)

(8cos(x)-4)(sin(x))=0,0<= x<2pi

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

sin (x) = 3511

(10)/(sin(30))= 3/(sin(b))

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan(θ/2-pi/6)=-1,0<= θ<2pi

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π