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

  • Thread starter seanistic
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In summary, the conversation discusses how to solve the equation sin(x)+2sin(x)cos(x)=0, with the given restrictions of 0<x<2(pie). The individual asking the question initially believes the equation cannot be solved without using identities, but is then reminded that the distributive law can be used to simplify the equation. The conversation then delves into the commutative property of multiplication and clarifies that sinx+2cosxsinx is equivalent to sinx+2sinxcosx. The conversation ends with the realization that the 2 in the equation can be moved around and does not have to stay with its respective function.
  • #1
seanistic
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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.
 
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  • #2
Why do you need to use identities? Use the distributive law; in other words, pull the sin(x) out.
 
  • #3
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.
 
  • #4
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
 
  • #5
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".
 
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  • #6
Yes, it is, since real number (and real function) multiplication is a commutative operation.
 
  • #7
seanistic said:
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
 
  • #8
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 = ...
 
  • #9
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?
 
  • #10
probably, that's why soemtiems it helps to write sinx,cosx out as sin(x), cos(x)
 

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

1. What is the solution to the equation?

The solution to the equation is x = π/6 or x = 5π/6.

2. How do you solve for x in this equation?

To solve for x, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x). Therefore, the equation can be rewritten as 0 < x < 2π sin(2x) = 0. From this, we can solve for x by setting sin(2x) = 0 and solving for x.

3. Can this equation have more than two solutions?

No, this equation can only have two solutions because the given interval is 0 < x < 2π, and sin(2x) has a period of π. Therefore, there can only be two distinct solutions within this interval.

4. Is there a way to check if the solutions are correct?

Yes, we can substitute the solutions back into the original equation and see if it satisfies the equation. If it does, then the solutions are correct.

5. Can this equation be solved using a graphing calculator?

Yes, this equation can be solved using a graphing calculator by graphing the function y = 2π sin(x)+2sin(x)cos(x) and finding the x-intercepts, which correspond to the solutions of the equation.

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