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受欢迎的三角函数 >

4cos^4(x)-1=0

解答

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

解答

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 = 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

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

用替代法求解

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

令

: cos (x) = u

4 u^{4} - 1 = 0

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

4 u^{4} - 1 = 0

将

1

到右边

4 u^{4} - 1 = 0

两边加上

1

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

化简

4 u^{4} = 1

4 u^{4} = 1

两边除以

4

4 u^{4} = 1

两边除以

4

\frac{4 u ^{4}}{4} = \frac{1}{4}

化简

u^{4} = \frac{1}{4}

u^{4} = \frac{1}{4}

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} = \frac{1}{4}

解

v^{2} = \frac{1}{4} : v = \frac{1}{4}, v = - \frac{1}{4}

v^{2} = \frac{1}{4}

对于

(g (x))^{2} = f (a)

解为

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

v = \frac{1}{4}, v = - \frac{1}{4}

v = \frac{1}{4}, v = - \frac{1}{4}

代回

v = u^{2}

，求解

u

解

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

u^{2} = \frac{1}{4}

化简

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{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

= \frac{1}{2}

使用法则

1 = 1

= \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

u^{2} = - \frac{1}{4} : u = i \frac{1}{2}, u = - i \frac{1}{2}

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

化简

- \frac{1}{4} : - \frac{1}{2}

- \frac{1}{4}

化简

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{1}{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

= \frac{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

对于

x^{2} = f (a)

解为

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

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

化简

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1}{2}

化简

- - \frac{1}{2} : - i \frac{1}{2}

- - \frac{1}{2}

化简

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

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

解为

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

u = cos (x)

代回

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

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

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

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

无解

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

无解

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

无解

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

无解

合并所有解

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 = 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

作图

流行的例子

arccos(2x)-arccos(x)= pi/3

arccos (2 x) - arccos (x) = \frac{π}{3}

cos(2θ)= 1/2 ,0<= θ<= 2pi

cos (2 θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

tan(x-pi/2)=(sqrt(3))/3

tan (x - \frac{π}{2}) = \frac{3}{3}

2 = 2 cos (3 x)

cos(2x)=sqrt(3)

cos (2 x) = 3