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]}