find domain of (4cosx)/(1-sinx)
To find the domain of the function, we need to determine the values of x that make the denominator equal to zero since division by zero is undefined.
The denominator of the function is 1 - sin(x). For the denominator to be zero, sin(x) must equal 1. This occurs when x = π/2 + 2πk, where k is an integer.
Therefore, the domain of the function is all real numbers except x = π/2 + 2πk for any integer k.
To find the domain of the function (4cosx)/(1-sinx), we need to consider the values of x for which the function is defined.
The function (4cosx)/(1-sinx) is defined for all values of x except for those that make the denominator equal to zero. So we need to find the values of x that satisfy the equation 1 - sinx = 0.
Solving this equation, we get sinx = 1.
The sin function takes values between -1 and 1, so there is no value of x for which sinx equals 1. Therefore, there are no values of x that make the denominator equal to zero, and the function (4cosx)/(1-sinx) is defined for all real numbers.
Hence, the domain of the function is the set of all real numbers (-∞, ∞).
To find the domain of the function \(f(x) = \frac{4\cos(x)}{1 - \sin(x)}\), we need to identify which values of \(x\) are not defined.
The function \(f(x)\) is not defined when the denominator is equal to zero because division by zero is undefined. So, we need to find the values of \(x\) for which \(\sin(x) = 1\), as this would make the denominator equal to zero.
The sine function, \(\sin(x)\), has a maximum value of 1. It occurs at two points within one period: \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). After \(x = \frac{3\pi}{2}\), the sine function starts decreasing again.
Therefore, the domain of the function \(f(x) = \frac{4\cos(x)}{1 - \sin(x)}\) is all real values of \(x\) except for \(x = \frac{\pi}{2} + n\pi\) and \(x = \frac{3\pi}{2} + n\pi\), where \(n\) is an integer.
In interval notation, the domain is \((- \infty, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \infty)\).