Solve $cos^2x+cosx=0$ ?

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Answer 1 · 2016-12-15T03:44:53

$x= pi/2 +- npi, pi +-n2pi$ , where $n$ is a natural number

Let's have $cosx=y$ , so we can write:

$cos^2x+cosx=0$

as

$y^2+y=0$

Solving for y:

$y(y+1)=0$

$y=0, y=-1$

Substituting back in:

$cosx=0, cosx=-1$

Now let's take them one at a time:

$cosx=0$

$x= pi/2 +- npi$

$cosx=-1$

$x=pi +-n2pi$

In both cases $n$ is a natural number.

And we can see these solutions in the graph of $cosx$

graph{cosx [-6.25, 6.25, -1.5, 1.5]}

Solve cos^2x+cosx=0cos^2x+cosx=0?