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Explain how solve this hard equation sin^3 x+sin^2 3x+sin x=3?
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Hence, evaluating the solution to the given equation yields
HOW DID YOU GET HERE? MAY YOU EXPLAIN?
Posted by oldnick on May 9, 2013 at 2:46 AM (Answer #2)
You need to remember that the sine function take on y values in interval [-1,1], hence, the following inequality holds for sine function, such that:
`-1 =< sin x <= 1 => -1 =< sin 3x <= 1`
Raising to square `sin 3x <= 1` yields:
`sin^2 3x <= 1`
Raising to cube `sin x <= 1` yields:
`sin^3 x <= 1`
Hence, evaluating the given relation yields:
`sin^3 x + sin^2 3x + sin x <= 1 + 1 + 1 = 3 => sin x = 1 => x = sin^(-1) 1 + n*pi => x = (-1)^k*pi/2 + n*pi` Hence, evaluating the solution to the given equation yields `x = (-1)^k*pi/2 + n*pi.`
Posted by sciencesolve on May 8, 2013 at 6:09 PM (Answer #1)
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