sin(2x) - sinx = 2cos^2 (x) - cosx Can someone show me the steps on how to find all the solutions in the interval [0,2pi)?
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Originally Posted by albo23 sin(2x) - sinx = 2cos^2 (x) - cosx Can someone show me the steps on how to find all the solutions in the interval [0,2pi)? edit ... pulled out the factor incorrectly. $#it happens.
Last edited by skeeter; Apr 25th 2012 at 01:56 PM. Reason: fixed the botched factoring job
Originally Posted by albo23 sin(2x) - sinx = 2cos^2 (x) - cosx Can someone show me the steps on how to find all the solutions in the interval [0,2pi)? Hi albo23, Slightly different approach. Use the double angle identity to replace Then, In which case, In the first case, In the second case, My answers differ from Skeeter's. Who is correct?
Last edited by masters; Apr 25th 2012 at 12:00 PM.
Originally Posted by masters My answers differ from Skeeter's. Who is correct? Skeeter got a bit carried away.
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