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July 1st, 2014, 01:21 AM   #1
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trigonometric

Hi all,
Can someone help me about below equations? What are the values of a and b?

2cos3a+2cos3b=1
2cos5a+2cos5b=1


Pleaze help me!
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July 1st, 2014, 12:11 PM   #2
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There may be a neater way. I would start by using the sum formulas to reduce every thing to cosa and cosb. You then have two equations in two unknowns.
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July 3rd, 2014, 09:07 AM   #3
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There is infinite number of solutions..... for it.

Let's assume there is solution for $\displaystyle 0 \le a,b \le \pi /2$

$\displaystyle b= \frac{\arccos ( \frac{1- 2 \cos (3a)}{2} )}{3} = \frac{\arccos ( \frac{1- 2 \cos (5a)}{2} )}{5}$

Now we have one variable and one equation.(?)

I conjecture that solution is near to 1.5.

Find x where f(x)=0 in $\displaystyle f(x) = \frac{\arccos ( \frac{1- 2 \cos (3x)}{2} )}{3} - \frac{\arccos ( \frac{1- 2 \cos (5x)}{2} )}{5}$
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