Solve for x: 0 < x < 2(pie) sin(x)+2sin(x)cos(x)=0

[seanistic]

solve for x, if 0< x < 2(pie)

sin(x)+2sin(x)cos(x)=0

Correct me if I am wrong but you can only solve this if the equation consists of all sin(x) or all cos(x). I realize that 2sinAcosA = sin2A but I am in section 1 which is "simple trigonometric equations" and section 2 is "using identities in trigonometric equations" so I don't see why they would put a problem that uses identities in sections 1.

 

[neutrino]

Why do you need to use identities? Use the distributive law; in other words, pull the sin(x) out.

 

[seanistic]

well I thought about that and I came up with sinx(1+2cosx) which I don't think is right because then I would get sinx+2cosxsinx.

 

[bob1182006]

when you have sinx(1+2cosx)=0 all you have to do is find the roots of the equation when sinx=0 and 1+2cosx=0

 

[seanistic]

I understand the part about finding the roots but I am confused that sinx+2sinxcosx = sinx(1+2cosx) because when I distribute i get sinx+2cosxsinx. is sinx+2cosxsinx the same as sinx+2sinxcosx?also I thought in order to solve for x you had to have the same trig function. Right now the equation would be equivilant to x+2xy=0 trying to solve for x. that's why I thought you had to have the indentity in order to have only one variable.

I know I am slow at this but in my defense I am teaching myself out of "Trigonometry - 5th edition by Charles P. Mckeague and Mark D. Turner".

 

[radou]

Yes, it is, since real number (and real function) multiplication is a commutative operation.

 

[neutrino]

is sinx+2cosxsinx the same as sinx+2sinxcosx?

Of course it is!

sin(x)(2+cos(x)) = 0
=> sin(x) = 0, cos(x) +0.5 = 0

 

[bob1182006]

yes sinx+2sinxcosx = sinx+2cosxsinx as long as you have the 2, cosx, and sinx being multiplied, it doesn't matter in what order you have them being multiplied, ex. sinx*cosx*2 = 2*cosx*sinx = cosx*2*sinx = ...

 

[seanistic]

ok, that makes sense now. I was under the impression that the 2 stayed with its function, perhaps I was thinking of cos2x or sin2x?

 

[bob1182006]

probably, that's why soemtiems it helps to write sinx,cosx out as sin(x), cos(x)