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September 28th, 2010, 03:01 AM  #1 
Newbie Joined: Sep 2010 Posts: 10 Thanks: 0  Find the solution set please help!
sqrt(3)sin x + cos x =1 Here's what I did, 1. square both sides 3sin^2 x + 2sqrt(3)sin x cos x + cos^2 x = 1 2. expand 3sin^2 x 33cos^2 x + 2sqrt(3)sin x cos x + cos^2 x = 1 3. combine 2cos^2 x + 2sqrt(3)sin x cos x = 2 4. extract 2cos x 2cos x (cos x  sqrt(3)sin x) = 2 so, => 2cos x = 2 => cos x = 1 => 0 deg. now I'm stopped here, => (cos x  sqrt(3)sin x) = 2 Please help! And addtl I just want to check my answer in this question, There is a bike with a pedal and gear measuring 10 radius and 7 radius respectively. If the bike's wheel(The one in line with the gear; rear wheel) is 42/pi in diameter, how many times the pedal will turn if the bike traveled 60 meters? Put the answer in rev. Thanks in advance 
September 28th, 2010, 05:12 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,491 Thanks: 889 Math Focus: Elementary mathematics and beyond  Re: Find the solution set please help!
?(3)sin(x) + cos(x) = 1, cos(x) = 0 is not a solution. Divide by cos(x): ?(3)tan(x) + 1 = sec(x) Square: 3tanē(x) + 2?(3)tan(x) + 1 = secē(x) = 1 + tanē(x) 2tanē(x) + 2?(3)tan(x) = 0 tanē(x) + ?(3)tan(x) = 0 tan(x)(tan(x) + ?(3)) = 0 ? tan(x) = 0, tan(x) = ?(3) ? x = 0, 2?/3. (? and 5?/3 are not solutions). 
September 29th, 2010, 04:58 AM  #3  
Global Moderator Joined: Dec 2006 Posts: 17,221 Thanks: 1294  Quote:
The original equation gives (?(3)/2)sin x + (1/2)cos x = 1/2, i.e., sin(x + ?/6) = sin(?/6) = sin(5?/6). Hence x = 2k? or (2/3)? + 2k?, where k is an integer. For the 2nd problem, in what units is the wheel's diameter given, and what working was done?  

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