What is the general solution of the following:
cosx (1+sinx/4 - 2cosx) + sinx (cosx/4-2sinx)=0
zeborg:
Is it (1+sinx)/4 or 1+sin(x/4) or 1+(sinx)/4 ?
1+sin(x/4). Sorry for the inconvenience caused :)
Gosh. Spent half an hour and still have no satisfying solution. The equation comes down to sum of two sine functions equal to 2, having theta (π/2 - x) and (5x/4), but that's only possible when they are both at their maximum value, i.e. π/2, which is apparently not possible in this case.
The answer given in my book is x=(4n+1)2 pi, n belongs to integers
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cos x(1+sin x/4 - 2cos x) + sin x(cos x/4 - 2sin x) = 0
L.H.S. :
cos x/4(1+sin x - 8cos x) + sin x/4(cos x - 8sin x)
=1/4(cos x(1+sin x - 8cos x) +sin x(cos x - 8sin x))
=1/4(cos x + cos x sin x - 8cos square x + sin x cos x - 8sin square x)
=1/4(sin x + cos x + 2 sin x cos x - 8(cos square x + sin square x))
=1/4(sin x + cos x + 2 sin x cos x - 8)
Now, 1/4(sin x + cos x + 2 sin x cos x - 8) = 0
=> sin x + cos x + 2 sin x cos x = 8
=> sin square x + cos square x + 2 sin square x cos square x = 64
=> 2 sin square x cos square x = 63
=> sin square x cos square x = 31.5
=> sin x cos x = root 31.5
=> sin x = root 31.5 / cos x
Hope this helps you... Please mark it as brainliest!!!
L.H.S. :
cos x/4(1+sin x - 8cos x) + sin x/4(cos x - 8sin x)
=1/4(cos x(1+sin x - 8cos x) +sin x(cos x - 8sin x))
=1/4(cos x + cos x sin x - 8cos square x + sin x cos x - 8sin square x)
=1/4(sin x + cos x + 2 sin x cos x - 8(cos square x + sin square x))
=1/4(sin x + cos x + 2 sin x cos x - 8)
Now, 1/4(sin x + cos x + 2 sin x cos x - 8) = 0
=> sin x + cos x + 2 sin x cos x = 8
=> sin square x + cos square x + 2 sin square x cos square x = 64
=> 2 sin square x cos square x = 63
=> sin square x cos square x = 31.5
=> sin x cos x = root 31.5
=> sin x = root 31.5 / cos x
Hope this helps you... Please mark it as brainliest!!!
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