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September 28th, 2010, 03:01 AM   #1
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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

3-3cos^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
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September 28th, 2010, 05:12 AM   #2
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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).
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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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