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Solve in the given interval of `-Pi <= x<= Pi` : `cos^2 2x + 2cos2x +1 = 0`

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sasuke24 | Student, Undergraduate | (Level 2) Honors

Posted May 15, 2013 at 10:11 PM via web

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Solve in the given interval of `-Pi <= x<= Pi` : `cos^2 2x + 2cos2x +1 = 0`

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oldnick | (Level 1) Valedictorian

Posted May 15, 2013 at 10:43 PM (Answer #1)

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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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llltkl | College Teacher | (Level 3) Valedictorian

Posted May 16, 2013 at 1:47 AM (Answer #2)

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`cos^(2)2x+2cos2x+1=0`          (given)

`rArr (cos2x)^2+2cos2x*1+(1)^2=0`

`=> (cos2x+1)^2=0`

`rArr cos2x+1=0`

`=> cos2x=-1=cos+-pi`

`rArr 2x=+-pi`

`rArr x=+-pi/2`

Therefore, the required solution for x is `+-pi/2` .

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