Help me solve this trig equation

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mynamesmurph

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sinx + 2sin(x/2) = cos(x/2) + 1

Maybe point me in the right direction?

Thanks folks!
 
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mynamesmurph said:
Replace x/2 with what? I'm not following.
Click to expand...

\(\displaystyle sin(x) = sin(2*(x/2)) = sin(2\theta)=2sin(\theta)cos(\theta)\)

So that
\(\displaystyle sinx + 2sin(x/2) = 2sin(\theta)cos(\theta)+2sin(x/2)=2sin(\theta)cos(\theta)+2sin(\theta)\)
\(\displaystyle cos(x/2) + 1 = cos(\theta)+1\)
and the original problem
sinx + 2sin(x/2) = cos(x/2) + 1
becomes
\(\displaystyle 2sin(\theta)cos(\theta)+2sin(\theta)=cos(\theta)+1\)

When you get the solution for \(\displaystyle \theta\), substitute back
\(\displaystyle \theta = \frac{x}{2}\)
 
2sin(θ)cos(θ)+2sin(θ)=cos(θ)+1

2sinΘ(cosΘ+1)=cosΘ+1

2sinΘ=1
sinΘ=1/2
(x/2)=π/6, 5π/6
sin(x/2)=π/3, 5π/3 within [0,2π)

I graph this and it is confirmed. Thank you.

------------------

I am a bit unclear on something though. Consider the following equation

sin2x=√3(sinx)

Initially, I solved it like this.

2sinxcosx=√3(sinx) **(dividing by sinx, which relates to my future question)**
2cosx=√3
cosx=√3/2
x=π/6, 11π/6 within [0,2π)

However, the graphing the two equations, I can see that points are correct, but I have 4 points of intersection. So, I went back and tinkered some more.

sin2x=√3(sinx)
2sinxcosx=√3(sinx)
2sinxcosx-√3(sinx)=0
sinx(2cosx-√3)=0
Now I have two factors equal to zero and I can solve and get my four points.
0, π, π/6, 11π/6 within [0,2π)

So I suppose, my question is what rule am I breaking when I divide my sinx? I realized I was doing this on a couple of problems and had to rework them to get all of the answers? Considering the first equation I posted, I divided both sides by (cosΘ+1) and eliminated those. I suppose I'm missing something.

Any help?
 
mynamesmurph said:
2sin(θ)cos(θ)+2sin(θ)=cos(θ)+1

2sinΘ(cosΘ+1)=cosΘ+1

2sinΘ(cosΘ+1)-(cosΘ+1) = 0

[2sin(Θ)-1][cos(Θ)+1] = 0

So

[cos(Θ)+1] = 0 → Θ = π → x = 2π (not within the given domain)

and

2sinΘ=1
sinΘ=1/2
(x/2)=π/6, 5π/6
sin(x/2)=π/3, 5π/3 within [0,2π)

I graph this and it is confirmed. Thank you.

------------------

I am a bit unclear on something though. Consider the following equation

sin2x=√3(sinx)

Initially, I solved it like this.

2sinxcosx=√3(sinx) **(dividing by sinx, which relates to my future question)**
2cosx=√3
cosx=√3/2
x=π/6, 11π/6 within [0,2π)

However, the graphing the two equations, I can see that points are correct, but I have 4 points of intersection. So, I went back and tinkered some more.

sin2x=√3(sinx)
2sinxcosx=√3(sinx)
2sinxcosx-√3(sinx)=0
sinx(2cosx-√3)=0
Now I have two factors equal to zero and I can solve and get my four points.
0, π, π/6, 11π/6 within [0,2π)

So I suppose, my question is what rule am I breaking when I divide my sinx? I realized I was doing this on a couple of problems and had to rework them to get all of the answers? Considering the first equation I posted, I divided both sides by (cosΘ+1) and eliminated those. I suppose I'm missing something.

Any help?
Click to expand...

When sin(x) = 0 is a possible solution, you cannot divide by sin(x) (effectively division by Zero - not allowed) to get rid of it.
 
sin2x=√3(sinx)
2sinxcosx=√3(sinx)


So what you are saying is because sinx might be zero, I cannot divide by it? Now, I'm trying to think how many times I may have divided a side by sin, cos, etc and not realized this. What is a good way to recognize when it may equal zero?
 
mynamesmurph said:
sin2x=√3(sinx)
2sinxcosx=√3(sinx)


So what you are saying is because sinx might be zero, I cannot divide by it? Now, I'm trying to think how many times I may have divided a side by sin, cos, etc and not realized this. What is a good way to recognize when it may equal zero?
Click to expand...

When you have an equation like
f(x) g(x) = a f(x)
then the solutions are
f(x) = 0
or, if f(x) \(\displaystyle \ne 0\), then
g(x) = a

In the cases here f(x) is sin(x) or cos(x) or ...
 
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