Solve for x the following 3cos(3x-1)=0.

Expert Answers
justaguide eNotes educator| Certified Educator

We have to solve 3*cos(3x - 1) = 0 for x.

3*cos(3x - 1) = 0

=> cos(3x - 1) = 0

=> 3x - 1 = arc cos 0

=> 3x - 1 = pi/2 + 2*n*pi and 3x - 1 = 3*pi/2 + 2*n*pi

=> 3x = pi/2 +1 + 2*n*pi and 3x = 3*pi/2 + 1 + 2*n*pi

=> x = pi/6 +1/3 + 2*n*pi/3 and x = pi/2 + 1/3 + 2*n*pi/3

The required solution is x = pi/6 +1/3 + 2*n*pi/3 and x = pi/2 + 1/3 + 2*n*pi/3

giorgiana1976 | Student

First, we'll divide by 3 both sides:

cos(3x-1)=0

Now, we'll take the inverse function both sides. For this reason, since the cosine function is not one to one function and inverse function can only be applied to one to one functions, we'll restrict the domain of cosine function to [0;pi].

3x - 1 = +/-arccos (0) + 2kpi, k is an integer number.

3x - 1= +/- pi/2  + 2kpi

We'll add 1 both sides:

3x = +/- pi/2  + 2kpi + 1

We'll divide by 3 to isolate x:

x = +/- pi/6 +  2kpi/3 + 1/3

The solutions of the equation over the range [0;pi] are:

x = pi/6 + 1/3

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