Solve the equation 2cos(3x) -1 = 0 for values in the interval 0=< x =< pi

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We have to solve the equation 2cos(3x) -1 for values in the interval 0=< x =< pi.

Now we see that 2cos(3x) -1 is not an equation, let's equate it to zero.

2cos(3x) -1 = 0

=> 2 cos 3x -1 = 0

=> 2 cos 3x = 1

=> cos 3x = 1/2

in the given t=interval we see that cos 60= 1/2

=> cos 3x = cos 60

=> 3x = 60

=> x = 20 degrees.

**Therefore the required result is x = 20 degrees.**

Given the equation 2cos(3x) -1 = 0

We need to solve for x if 0=< x =< pi.

==> 2cos(3x) -1 = 0

We will add 1 to both sides.

==> 2cos(3x) = 1

Now we will divide by 2.

==> cos(3x) = 1/2

Then we know that:

3x = pi/3

Now we will divide by 3.

**==> x = pi/9 **

To solve for x : 2cos(3x)-1 = 0.

x should be in (0, pi) interval.

Solution:

2cos3x-1 = 0.

=> 2cos3x = 1.

cos3x = 1/2.

3x = 2npi+pi/3, or 3x= 2npi-pi/3.

=> 3x = pi/3, for n = 0.

So x = pi/9 which is a solution in 0 < x< = pi.

When n=1, 3x = 2pi+pi/3 , or x= 2pi-pi/3.

So x = 7p/9 , or x= 5pi/9.

Therefore x = {pi/9, 5pi/3 and 7pi/3} are the solutions in (0, pi) interval.

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