How do you solve for x in $cos 2 x + 2 cos x + 1 = 0$ $cos2x+2cosx+1=0$ ?

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How do you solve for x in $cos 2 x + 2 cos x + 1 = 0$ $cos2x+2cosx+1=0$ ?

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Answer 1 · 2015-02-13T21:29:04

$cos (2 x) + 2 cos (x) + 1 = 0$ $cos(2x)+2cos(x)+1 = 0$ when
$x = \frac{π}{2}$ $x = pi/2$ (plus $n π$ $n pi$ for all integer $n$ $n$ )
or
$x = π$ $x = pi$ (plus $2 n π$ $2n pi$ for all integer $n$ $n$ )

How we got this :
$cos (2 x) + 2 cos (x) + 1 = 0$ $cos(2x) + 2cos(x) + 1 = 0$
$2 {cos}^{2} (x) - 1 + 2 cos (x) + 1 = 0$ $2cos^2(x) - 1 + 2cos(x )+ 1 = 0$ by a double angle formula for $cos (2 x)$ $cos(2x)$

$(2) ({cos}^{2} (x) + cos (x)) = 0$ $(2) (cos^2(x) + cos(x) ) = 0$
$(2) (cos (x)) (cos (x) + 1) = 0$ $(2) (cos(x)) (cos(x)+1) = 0$

So
$cos (x) = 0$ $cos(x) = 0$ which gives $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}$ $x = pi/2, (3pi)/2, (5pi)/2$ etc.
or
$cos (x) = - 1$ $cos(x) = -1$ which gives $x = π, 3 π, 5 π$ $x = pi, 3pi, 5pi$ etc.

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How do you solve for x in $cos 2 x + 2 cos x + 1 = 0$ $cos2x+2cosx+1=0$ ?

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