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**lachtach** · 09-25-2010, 12:57 AM

GPT: $\frac{1}{sinx } + \frac{1}{sin(x-\frac{3.\pi}{2})} = 4sin( \frac{7.\pi}{4} - x)$

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**kuta tutu** · 09-25-2010, 01:19 AM

Nguyên văn bởi lachtach

GPT: $\frac{1}{sinx } + \frac{1}{sin(x-\frac{3.\pi}{2})} = 4sin( \frac{7.\pi}{4} - x)$

ta có : $sin(x-\frac{3.\pi}{2}) = sin(x+\frac{\pi}{2}) = cosx$

$sin(\frac{7.\pi}{4}-x) = sin(\frac{-\pi}{4}-x) = -sin({x+\frac{\pi}{4})$

nên $PT <=> \frac{1}{sinx} + \frac{1}{cosx} = -4sin({x+\frac{\pi}{4})$ $dk : x$ khác $\frac{k\pi}{2}$

$<=> sinx+cosx = -4sinxcosxsin(x+\frac{\pi}{4})$

$<=> sinx + cosx = -4sinxcosx(sinx+cosx).\frac{1}{\sqrt{2}}$

$<=> sinx + cosx = o$ hoặc $sin2x = \frac{-1}{\sqrt{2}}$

$=> x = \frac{-\pi}{4} + k\pi$ hoặc $x = \frac{-\pi}{8} = k\pi$ hoặc $x = \frac{5\pi}{8} + k\pi$ ( k thuộc Z) .

**kuta tutu** · 09-25-2010, 01:19 AM

Nguyên văn bởi lachtach

GPT: $\frac{1}{sinx } + \frac{1}{sin(x-\frac{3.\pi}{2})} = 4sin( \frac{7.\pi}{4} - x)$

ta có : $sin(x-\frac{3.\pi}{2}) = sin(x+\frac{\pi}{2})= cosx$

$sin(\frac{7.\pi}{4}-x) = sin(\frac{-\pi}{4}-x) = - sin({x+\frac{\pi}{4})$

nên $PT <=> \frac{1}{sinx} + \frac{1}{cosx} = -4.sin({x+\frac{\pi}{4})$ $dk : x$ khác $\frac{k\pi}{2}$

$<=> sinx+cosx = -4sinxcosxsin(x+\frac{\pi}{4})$

$<=> sinx + cosx = -4sinxcosx(sinx+cosx).\frac{1}{\sqrt{2}}$

$<=> sinx + cosx = o$ hoặc $sin2x = \frac{-1}{\sqrt{2}}$

$=> x = \frac{-\pi}{4} + k\pi$ hoặc $x = \frac{-\pi}{8} = k\pi$ hoặc $x = \frac{5\pi}{8} + k\pi$ ( k thuộc Z) .