Can somebody please help me find the domain of:
1 / cos4x
If there are any working out, could you be kind enough to show your steps? Thankyou. All help is of course, appreciated.
1/cos4x is real as long as cos4x is not equal to zero.
the period of cos4x is 2pi/4 or pi/2 (90º)
so the first time that cos4x is zero is at x=pi/8
repeatedly adding or subtracting pi/4 (45º) to this will give us another zero.
so the domain for 1/cos4x is
x is any real number except x = pi/8 ± kpi/4 where k is an integer.
Of course y = 1/cos4x is the same as
y = sec 4x, which would have vertical asymptotes at the values of x described above.
To find the domain of the function 1 / cos(4x), we need to consider the values of the input variable (x) for which the function is defined.
The cosine function, cos(x), is defined for all real numbers. However, the function 1 / cos(4x) can have restrictions on its domain due to division by zero.
In this case, we need to determine the values of x for which cos(4x) is equal to zero because dividing by zero is undefined.
The values where cos(4x) equals zero can be found by solving the equation cos(4x) = 0.
To solve this equation, we set the argument (4x) of the cosine function equal to the values that make the cosine function zero. The cosine function is zero at every odd multiple of π/2.
So, we set 4x equal to (2n+1)(π/2), where n is an integer.
Simplifying the equation:
4x = (2n+1)(π/2)
x = [(2n+1)(π/2)] / 4
x = (2n+1)(π/8)
The solution x = (2n+1)(π/8) represents all the values of x for which the cosine function is zero.
Now, the domain is the set of real numbers except the values of x when cos(4x) is equal to zero. Thus, we can state the domain as:
Domain: x ≠ (2n+1)(π/8) for all integers n.
This means that x can take any real value except for the values given by the formula (2n+1)(π/8).
Remember that this explanation assumes that the variable x represents real numbers. If x represents complex numbers, the domain may be larger.