July 1st, 2014, 01:21 AM  #1 
Newbie Joined: Jul 2014 From: london Posts: 1 Thanks: 0  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! 
July 1st, 2014, 12:11 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,804 Thanks: 715 
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.

July 3rd, 2014, 09:07 AM  #3 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
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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