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受欢迎的三角函数 >

((cos^3(a)))/((2cos^2(a)-1))=cos(a)

解答

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

解答

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

+1

度数

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

求解步骤

\frac{( cos ^{3} ( a ) )}{( 2 cos ^{2} ( a ) - 1 )} = cos (a)

用替代法求解

\frac{cos ^{3} ( a )}{2 cos ^{2} ( a ) - 1} = cos (a)

令

: cos (a) = u

\frac{u ^{3}}{2 u ^{2} - 1} = u

\frac{u ^{3}}{2 u ^{2} - 1} = u : u = 0, u = - 1, u = 1

\frac{u ^{3}}{2 u ^{2} - 1} = u

在两边乘以

2 u^{2} - 1

\frac{u ^{3}}{2 u ^{2} - 1} = u

在两边乘以

2 u^{2} - 1

\frac{u ^{3}}{2 u ^{2} - 1} (2 u^{2} - 1) = u (2 u^{2} - 1)

化简

u^{3} = u (2 u^{2} - 1)

u^{3} = u (2 u^{2} - 1)

解

u^{3} = u (2 u^{2} - 1) : u = 0, u = - 1, u = 1

u^{3} = u (2 u^{2} - 1)

展开

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

u (2 u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = u, b = 2 u^{2}, c = 1

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

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

化简

2 u^{2} u - 1 \cdot u : 2 u^{3} - u

2 u^{2} u - 1 \cdot u

2 u^{2} u = 2 u^{3}

2 u^{2} u

使用指数法则:

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

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

= 2 u^{2 + 1}

数字相加:

2 + 1 = 3

= 2 u^{3}

1 \cdot u = u

1 \cdot u

乘以:

1 \cdot u = u

= u

= 2 u^{3} - u

= 2 u^{3} - u

u^{3} = 2 u^{3} - u

交换两边

2 u^{3} - u = u^{3}

将

u^{3}

para o lado esquerdo

2 u^{3} - u = u^{3}

两边减去

u^{3}

2 u^{3} - u - u^{3} = u^{3} - u^{3}

化简

u^{3} - u = 0

u^{3} - u = 0

因式分解

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

因式分解出通项

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

u^{3} - u

使用指数法则:

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

u^{3} = u^{2} u

= u^{2} u - u

因式分解出通项

u

= u (u^{2} - 1)

= u (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 (u + 1) (u - 1)

u (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 + 1 = 0 or u - 1 = 0

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 = - 1, u = 1

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

验证解

找到无定义的点（奇点）:

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

取

\frac{u ^{3}}{2 u ^{2} - 1}

的分母，令其等于零

解

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

2 u^{2} - 1 = 0

将

1

到右边

2 u^{2} - 1 = 0

两边加上

1

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

化简

2 u^{2} = 1

2 u^{2} = 1

两边除以

2

2 u^{2} = 1

两边除以

2

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

化简

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

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

对于

x^{2} = f (a)

解为

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

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

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

\frac{1}{2}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{2}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{2}

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

- \frac{1}{2}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{2}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{2}

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

以下点无定义

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

将不在定义域的点与解相综合:

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

u = cos (a)

代回

cos (a) = 0, cos (a) = - 1, cos (a) = 1

cos (a) = 0, cos (a) = - 1, cos (a) = 1

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

cos (a) = 0

cos (a) = 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}

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

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

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

cos (a) = - 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}

a = π + 2 πn

a = π + 2 πn

cos (a) = 1 : a = 2 πn

cos (a) = 1

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

合并所有解

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

作图

流行的例子

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