Solve the equation 2cos(3x) -1 = 0 for values in the interval 0=< x =< pi
3 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
hala718 | High School Teacher | (Level 1) Educator Emeritus
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
neela | High School Teacher | (Level 3) Valedictorian
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.