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

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### 2 Answers

`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` .

**Sources:**

`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`