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hoanghondo94

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Câu 1:

[TEX]\Leftrightarrow 2\sqrt{2}cos(\frac{\pi}{4}+\frac{\pi}{6}-x)sinx=1 \\\\ \Leftrightarrow2\sqrt{2}[\frac{1}{\sqrt{2}}cos(\frac{\pi}{6}-x) -\frac{1}{\sqrt{2}}sin(\frac{\pi}{6}-x)]sinx=1 \\\\ \Leftrightarrow2(cos(\frac{\pi}{6}-x) -sin(\frac{\pi}{6}-x)]sinx=1 \\\\ \Leftrightarrow2[\frac{\sqrt{3}}{2}cosx +\frac{1}{2}sinx -\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx]sinx=1 \\\\ \Leftrightarrow2\sqrt{3}sinxcosx=1 \\\\\Leftrightarrow sin2x=\frac{1}{\sqrt{3}} \\\\\Leftrightarrow\left[\begin{ x= \frac{1}{2}arcsin\frac{1}{\sqrt{3}}+k\pi \\ x=\frac{\pi}{2} -\frac{1}{2}arcsin\frac{1}{\sqrt{3}}+k\pi[/TEX]

Câu 2:

$$(1)\Leftrightarrow 2cosx+\frac{1}{3}cos^2x=\frac{8}{3}+sin2x+3sinx+\frac{1}{3 }sin^2x$$
$$\Leftrightarrow 2cosx+\frac{1}{3}cos2x-\frac{8}{3}-sin2x-3sinx=0$$
$$\Leftrightarrow -2sin^2x+9sinx-7+6cosx-3sin2x=0$$
$$\Leftrightarrow-2(sinx-1)(sinx-\frac{7}{2})+6cosx(1-sinx)=0$$.
 
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