SOLUTION: Solve for x in [0, 2pi]: cosx-sinx=-1

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Question 678143: Solve for x in [0, 2pi]: cosx-sinx=-1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
cos%28x%29-sin%28x%29=-1
To solve this equation we will be using the (less frequently used) identity:
A%2Acos%28x%29+%2B+B%2Asin%28x%29+=+C%2Acos%28x-Q%29
where
C+=+sqrt%28A%5E2%2BB%5E2%29
cos%28Q%29+=+A%2FC
sin%28Q%29+=+B%2FC

In your equation A = 1 and B = -1 (because of the subtraction). So:
C+=+sqrt%28%281%29%5E2%2B%28-1%29%5E2%29+=+sqrt%281%2B1%29+=+sqrt%282%29
cos%28Q%29+=+1%2Fsqrt%282%29+=+sqrt%282%29%2F2
sin%28Q%29+=+%28-1%29%2Fsqrt%282%29+=+-sqrt%282%29%2F2

From the last two equations we should find that Q+=+-pi%2F4. Using the identity and the values we found for C and Q, we can replace the left side of your equation:
sqrt%282%29%2Acos%28x-%28-pi%2F4%29%29+=+-1
which simplifies to:
sqrt%282%29%2Acos%28x%2Bpi%2F4%29+=+-1

With the equation in this form we can now find the general solution for x. First we divide by sqrt%282%29
cos%28x%2Bpi%2F4%29+=+-1%2Fsqrt%282%29
which simplifies to:
cos%28x%2Bpi%2F4%29+=+-sqrt%282%29%2F2
We should recognize that -sqrt%282%29%2F2 is a special angle value for cos. We should recognize that the reference angle will be pi%2F4 and that cos's will be negative in the 2nd and 3rd quadrants. So our general solution will be:
x%2Bpi%2F4+=+pi-pi%2F4+%2B+2pi%2An (for the 2nd quadrant angles)
x%2Bpi%2F4+=+pi%2Bpi%2F4+%2B+2pi%2An (for the 3rd quadrant angles)
where "n" represents any integer. (NOTE: Some books.teachers uses "k" instead of "n". The letter used does not matter. What matters is that it represents an integer.) These equations simplify to:
x%2Bpi%2F4+=+3pi%2F4+%2B+2pi%2An (for the 2nd quadrant angles)
x%2Bpi%2F4+=+5pi%2F4+%2B+2pi%2An (for the 3rd quadrant angles)
Subtracting pi%2F4 from each side we get:
x+=+2pi%2F4+%2B+2pi%2An (for the 2nd quadrant angles)
x+=+4pi%2F4+%2B+2pi%2An (for the 3rd quadrant angles)
which simplify to:
x+=+pi%2F2+%2B+2pi%2An (for the 2nd quadrant angles)
x+=+pi+%2B+2pi%2An (for the 3rd quadrant angles)
This is the general solution. (The general solution expresses the infinite set of solutions to an equation.)

Often problems request a specific solution. For example: "Find the least positive solution to ..." or "Find all solutions to ... between 0 and 2pi". Your problem asks for solutions in the interval [0, 2pi]. To find these specific solution(s) you try different integers for "n" until you're satisfied that you have found all possible specific solutions. Trying different n's each equation of our general solution. For the first equation:
x+=+pi%2F2+%2B+2pi%2An (for the 2nd quadrant angles)
If n = 0 then x+=+pi%2F2
If n = 1 (or any higher integer) then x is more than 2pi.
If n is negative then x is negative (which is below 0).

For the second equation:
x+=+pi+%2B+2pi%2An (for the 3rd quadrant angles)
If n = 0 then x+=+pi
If n = 1 (or any higher integer) then x is more than 2pi.
If n is negative then x is negative (which is below 0).

So the only solutions to your equation that are in the interval [0, 2pi] are: x+=+pi%2F2 and x+=+pi.

P.S. Just because n values of 0 were the only n values that gave us solutions in the desired interval, do not assume
  • n = 0 will always give you a solution you want. n = 0 will sometimes not give you a desired solution.
  • n values that are not 0 will fail to give you solutions you desire. Sometimes other n values will give desired solutions.