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How to solve the equation cos(2x+ pi/2)=cos(x- pi/2)?
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Best answer as selected by question asker.
There are, at least, 2 methods of solving the equation given.
First, let's solve the equation as a general one:
2x+ pi/2 = +/- arccos(cos(x- pi/2)) + 2*k*pi
Let's find the positive solution:
2x+ pi/2 = x- pi/2 + 2*k*pi
We'll move the unknown to the left side:
2x-x=-pi/2 - pi/2 + 2*k*pi
x=-2*pi/2 + 2*k*pi
x=-pi + 2*k*pi
Let's find now the negative solution:
2x+ pi/2 = -x + pi/2 + 2*k*pi
After moving the unknown to the left and simplifying the similar terms, we'll have:
2x+x = 2*k*pi
3x = 2*k*pi
x = 2*k*pi/3
The solutions of the equation are:
Another manner of solving would be to transform the difference of functions into a product, using the formula:
cos a-cos b= 2*sin(a+b)/2*sin(b-a)/2
Posted by giorgiana1976 on April 13, 2010 at 1:45 AM (Answer #1)
If cos(A) = cos(B), then
case 1: A = n*2Pi+B, where n is an integer,
case2: A = n*2Pi-B
A = 2x+pi/2, B = x-pi/2
2x+pi/2 = n*2pi+x-pi/2
x + pi = n*2pi
x = pi(2n-1), or x = -pi for n=0
2x+pi/2 = n*2pi-(x-pi/2)
2x+pi/2 = n*2pi-x+pi/2
3x = n*2pi
x = 2nPi/3, or x = 0 for n=0
Posted by kjcdb8er on April 13, 2010 at 2:01 AM (Answer #2)
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