How can i solve 4sin(x)cos(x) + cos(x) = 2

L

Lez

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Aug 12, 2017
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Hi All

I wonder if you can help me to solve:-

4sin(x)cos(x) + cos(x) = 2

I've tried tangent half angle substitutions for sin(x) and cos(x) & also substituting using sin^2(x) + cos^2(x) = 1 but it all gets very complicated & messy.

I've also tried converting to 2sin(2x) + cos(x) = 2 but can't get any farther.

Is there something simple I'm missing?

Many thanks
Lez :(
 
Why not just factor it?

What would you do with 4xy + y = 2?

That may lead somewhere. It may also do no good.

What's wrong with "complicated and messy"?
 
Last edited:
Many thanks for your reply

I did try 4xy + y = 2 but didn't get anywhere. I'll have another look though, as it was late in the day when i tried this approach.

"Complicated and messy" referred to attempts like:-

-3tan^4(x/2) - 4tan^3(x/2) - 4tan^2(x/2) + 4tan(x/2) - 1 =0

There were similar outcomes when other substitutions were tried.

Much appreciated.
Lez
 
I understand what "complicated and messy" means. You didn't answer my question.

1) Are you sure there is a solution? Prove it! Consider the function f(x) = \(\displaystyle 4\sin(x)\cos(x) + \cos(x) - 2\) Try x = \(\displaystyle \pi/3\) and x = \(\displaystyle \pi/2\). Does that prove anything?

2) Can you narrow down a solution to a relatively small Domain? Consider f(x) again. Try x = 0.27 and x = 0.28. Does that prove anything?

3) How sure are you that there is clean, algebraic solution? It's likely there is not and numerical methods will be required. Consider f(x) again. Try x = 0.272519916206. Does that suggest anything?

4) Don't quit too soon. These periodic function are likely to have periodic solutions. Try x = 1.1212742253189.

5) Think about the period of the function f(x). sin(2x) has what period? cos(x) has what period? Does this suggest where you should look for the next solution?

 
tkhunny said:
I understand what "complicated and messy" means. You didn't answer my question.

1) Are you sure there is a solution? Prove it! Consider the function f(x) = \(\displaystyle 4\sin(x)\cos(x) + \cos(x) - 2\) Try x = \(\displaystyle \pi/3\) and x = \(\displaystyle \pi/2\). Does that prove anything?

2) Can you narrow down a solution to a relatively small Domain? Consider f(x) again. Try x = 0.27 and x = 0.28. Does that prove anything?

3) How sure are you that there is clean, algebraic solution? It's likely there is not and numerical methods will be required. Consider f(x) again. Try x = 0.272519916206. Does that suggest anything?

4) Don't quit too soon. These periodic function are likely to have periodic solutions. Try x = 1.1212742253189.

5) Think about the period of the function f(x). sin(2x) has what period? cos(x) has what period? Does this suggest where you should look for the next solution?

Click to expand...

Apologies for not getting back sooner, as I was away. In answer to your questions:-

1) I was unable to prove there was a solution or solutions. I see from your example how to use trial and error for a numerical solution and how one can home in on a particular solution.

2) Solution is clearly between 0.27 and 0.28.

3) You're quite right - there is not necessarily a 'clean' algebraic solution. x = 0.272519916206 is clearly a solution as f(0.272519916206) = 0

4) Similarly x = 1.1212742253189 is a solution as f(1.1212742253189) = 0.

5) sin(2x) has a period of pi, whereas cos(x) has a period 2 x pi. Clearly further solutions will exist at the above mentioned solutions plus n.(2 x pi) where n is an integer. This can also be seen by plotting 2sin(2x) and cos(x) as separate functions to see where the repeats occur.

Many thanks for getting back on this. I appreciate your input and as a result I have certainly learned how to think about this problem in different ways, especially in terms of defining the domain of possible solutions. It's also made me aware of plotting individual terms of the equation to get a idea where potential solutions might exist.

Kind regards
Lez

 
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