how to solve cos(3x)=-1 on the interval [0,2pi]
part 1 : solve this equation for 3x, finidng all values on the interval [0,2pi] that satisfy the eqaution
part 2: use algebra to find all values of x between 0 and 2pi that satisfy the eqaution
You need to remember how to solve a simple trigonometric equation, such that:
`cos x = y => x = +-cos^(-1)y + 2n*pi `
Notice that the problem specifies that the solution needs to be in the interval `[0,2pi], ` hence, the general solution to the given equation is the following, such that:
`x = +-cos^(-1)y`
You need to remember that the cosine function is negative in quadrants 2 and 3 such that:
`cos 3x = -1 => 3x = cos^(-1)(-1) => 3x = pi => x = pi/3`
Since the cosine function is minimum at `x = pi` , over the interval `[0,2pi], ` hence, the problem has one solution, `x = pi/3` .
Hence, evaluating the solution to the given equation `cos 3x = -1` , over the interval `[0,2pi], ` yields `x = pi/3` .