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cos^4(x)= 1/16

解答

cos^{4} (x) = \frac{1}{16}

解答

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

cos^{4} (x) = \frac{1}{16}

用替代法求解

cos^{4} (x) = \frac{1}{16}

令

: cos (x) = u

u^{4} = \frac{1}{16}

u^{4} = \frac{1}{16} : u = \frac{1}{2}, u = - \frac{1}{2}, u = i \frac{1}{2}, u = - i \frac{1}{2}

u^{4} = \frac{1}{16}

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} = \frac{1}{16}

解

v^{2} = \frac{1}{16} : v = \frac{1}{16}, v = - \frac{1}{16}

v^{2} = \frac{1}{16}

对于

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

解为

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

v = \frac{1}{16}, v = - \frac{1}{16}

v = \frac{1}{16}, v = - \frac{1}{16}

代回

v = u^{2}

，求解

u

解

u^{2} = \frac{1}{16} : u = \frac{1}{2}, u = - \frac{1}{2}

u^{2} = \frac{1}{16}

化简

\frac{1}{16} : \frac{1}{4}

\frac{1}{16}

使用根式运算法则:

假定

a \geq 0, b \geq 0

= \frac{1}{16}

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

4^{2} = 4

= 4

= \frac{1}{4}

使用法则

1 = 1

= \frac{1}{4}

对于

x^{2} = f (a)

解为

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

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

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

\frac{1}{4}

使用根式运算法则:

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:

2^{2} = 2

= 2

= \frac{1}{2}

使用法则

1 = 1

= \frac{1}{2}

- \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:

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:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

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

解

u^{2} = - \frac{1}{16} : u = i \frac{1}{2}, u = - i \frac{1}{2}

u^{2} = - \frac{1}{16}

化简

- \frac{1}{16} : - \frac{1}{4}

- \frac{1}{16}

化简

\frac{1}{16} : \frac{1}{4}

\frac{1}{16}

使用根式运算法则:

假定

a \geq 0, b \geq 0

= \frac{1}{16}

16 = 4

16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

4^{2} = 4

= 4

= \frac{1}{4}

= - \frac{1}{4}

使用法则

1 = 1

= - \frac{1}{4}

对于

x^{2} = f (a)

解为

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

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

化简

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

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

使用根式运算法则:

假定

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \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:

2^{2} = 2

= 2

= i \frac{1}{2}

使用法则

1 = 1

= i \frac{1}{2}

将

i \frac{1}{2}

改写成标准复数形式:

\frac{1}{2} i

i \frac{1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

化简

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

- - \frac{1}{4}

化简

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

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

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

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

使用根式运算法则:

假定

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \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:

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2} i

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

流行的例子

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tan (2 θ) = \frac{500}{350 - 75}

sec (a) = - 2

tan^2(γ)+1=cos^2(γ)

tan^{2} (γ) + 1 = cos^{2} (γ)

sin^2(x)+8sin(x)-9=0

sin^{2} (x) + 8 sin (x) - 9 = 0