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Solve in the given interval of `-Pi <= x<= Pi` : `cos^2 2x + 2cos2x +1 = 0`
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`cos^2 2x+2cos2x+1=0` is a perfect square.
`(cos2x+1)^2=0` `cos2x+1=0` `cos2x=-1`
`2x=+- pi` then `x=+-pi/2`
Posted by oldnick on May 15, 2013 at 10:43 PM (Answer #1)
Therefore, the required solution for x is `+-pi/2` .
Posted by llltkl on May 16, 2013 at 1:47 AM (Answer #2)
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