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受欢迎的三角函数 >

r=(2a)/((1+cos(x)))

解答

r = \frac{2 a}{( 1 + cos ( x ) )}

解答

x = arccos (\frac{2 a}{r} - 1) + 2 πn, x = - arccos (\frac{2 a}{r} - 1) + 2 πn

求解步骤

r = \frac{2 a}{( 1 + cos ( x ) )}

交换两边

\frac{2 a}{1 + cos ( x )} = r

在两边乘以

1 + cos (x)

\frac{2 a}{1 + cos ( x )} = r

在两边乘以

1 + cos (x)

\frac{2 a}{1 + cos ( x )} (1 + cos (x)) = r (1 + cos (x))

化简

2 a = r (1 + cos (x))

2 a = r (1 + cos (x))

交换两边

r (1 + cos (x)) = 2 a

两边除以

r; r \neq = 0

r (1 + cos (x)) = 2 a

两边除以

r; r \neq = 0

\frac{r ( 1 + cos ( x ) )}{r} = \frac{2 a}{r}; r \neq = 0

化简

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

将

1

到右边

1 + cos (x) = \frac{2 a}{r}; r \neq = 0

两边减去

1

1 + cos (x) - 1 = \frac{2 a}{r} - 1; r \neq = 0

化简

cos (x) = \frac{2 a}{r} - 1; r \neq = 0

cos (x) = \frac{2 a}{r} - 1; r \neq = 0

使用反三角函数性质

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

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

的通解

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

x = arccos (\frac{2 a}{r} - 1) + 2 πn, x = - arccos (\frac{2 a}{r} - 1) + 2 πn

x = arccos (\frac{2 a}{r} - 1) + 2 πn, x = - arccos (\frac{2 a}{r} - 1) + 2 πn

作图

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