cos^2(x)+cos(x)sin(x)=0 solve for x

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cos^2(x)+cos(x)sin(x)=0 solve for x

Mathematics
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Try factoring
I factored a cos(x) out but then im stuck
cos(x)( ? + ? ) =0 So what should be in the places of those ?'s once you factor out cos(x)?

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Other answers:

cos(x)(cosx+sinx)
Yes. Set both factors equal to 0. cos(x)=0 ; cos(x)+sin(x)=0
Solve cos(x)=0 for x And find the value x so that cos(x) and sin(x) are opposites (i.e cos(x)=-sin(x))
cosx=pie/2 ?
cos(x)=0 for x=pi/2 cos(x) does not equal pi/2 But one value that satisfies cos(x)=0 is x=pi/2 Are there any other values?
x=3pi/2
Ok and you can also say we have x=pi/2+2npi where n is an integer or x=3pi/2+2n pi where n is an integer
We also need to find the values of x where cos(x) and sin(x) have opposite values.
|dw:1357920385873:dw| As you can see they will be opposite in quadrant 2 and 4.
|dw:1357920430907:dw| These points is where cos(x) and sin(x) are opposite. Now what angle are these points at from the initial ray?
135 and 315?
Correct.
Now should your answer be in radians or degrees?
um radians?
So 135*pi/180 or 315*pi/180
Don't forget we can go around the circle infinitely many times from those angles.
So 135pi/180+2npi , 315pi/180+2npi
Don't forget the other solutions we wrote above. Good job. :)
thanks!

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