Algebra Pre-Algebra and Basic Algebra Math Forum

 September 28th, 2010, 05:12 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 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
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Quote:
 Originally Posted by hirano 4. extract -2cos x -2cos x (cos x - sqrt(3)sin x) = -2 so, => -2cos x = -2
The above deduction was incorrect, since it was sqrt(3)sin x + cos x that was 1, not cos x - sqrt(3)sin x.

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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