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`

So :

`2x=+- pi`     then    `x=+-pi/2`

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