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受欢迎的三角函数 >

cos^5(x)=cos(x)

解答

cos^{5} (x) = cos (x)

解答

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

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

cos^{5} (x) = cos (x)

用替代法求解

cos^{5} (x) = cos (x)

令

: cos (x) = u

u^{5} = u

u^{5} = u : u = 0, u = i, u = - i, u = - 1, u = 1

u^{5} = u

将

u

para o lado esquerdo

u^{5} = u

两边减去

u

u^{5} - u = u - u

化简

u^{5} - u = 0

u^{5} - u = 0

因式分解

u^{5} - u : u (u^{2} + 1) (u + 1) (u - 1)

u^{5} - u

因式分解出通项

u : u (u^{4} - 1)

u^{5} - u

使用指数法则:

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

u^{5} = u^{4} u

= u^{4} u - u

因式分解出通项

u

= u (u^{4} - 1)

= u (u^{4} - 1)

分解

u^{4} - 1 : (u^{2} + 1) (u + 1) (u - 1)

u^{4} - 1

将

u^{4} - 1

改写为

(u^{2})^{2} - 1^{2}

u^{4} - 1

将

1

改写为

1^{2}

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

使用指数法则:

a^{b c} = (a^{b})^{c}

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

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(u^{2})^{2} - 1^{2} = (u^{2} + 1) (u^{2} - 1)

= (u^{2} + 1) (u^{2} - 1)

分解

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

将

1

改写为

1^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= (u^{2} + 1) (u + 1) (u - 1)

= u (u^{2} + 1) (u + 1) (u - 1)

u (u^{2} + 1) (u + 1) (u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or u^{2} + 1 = 0 or u + 1 = 0 or u - 1 = 0

解

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

将

1

到右边

u^{2} + 1 = 0

两边减去

1

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

化简

u^{2} = - 1

u^{2} = - 1

对于

x^{2} = f (a)

解为

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

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解为

u = 0, u = i, u = - i, u = - 1, u = 1

u = cos (x)

代回

cos (x) = 0, cos (x) = i, cos (x) = - i, cos (x) = - 1, cos (x) = 1

cos (x) = 0, cos (x) = i, cos (x) = - i, cos (x) = - 1, cos (x) = 1

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

cos (x) = i :

无解

cos (x) = i

无解

cos (x) = - i :

无解

cos (x) = - i

无解

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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

合并所有解

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

作图

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sin^{22} (x) = \frac{1}{4}

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