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Question

$2\sqrt{3}sinx(1+cosx)-4cosx\sin^{2}\frac{x}{2} =3$

GreenmjlkTea FeelingTea · Answer 1 · 5/25/2013 3:32:02 PM

Áp dụng công thức góc nhân đôi ta có

$2sin^2{\frac{x}{2}}=1-cosx$ . Thay vào phương trình đã cho ta có

$2\sqrt{3}sinx(1+cosx)-2cosx(1-cosx)=3$

$\Leftrightarrow (2\sqrt3sinx-2cosx)+(2\sqrt3sinx.cosx+2cos^2x)=3$

$\Leftrightarrow (2\sqrt3sinx-2cosx)+(2\sqrt3sinx.cosx+cos2x+1)=3$

$\Leftrightarrow(2\sqrt3sinx-2cosx)+(\sqrt3sin2x+cos2x)=2 (1)$

Chú ý rằng

$2\sqrt3sinx-2cosx=4(\frac{\sqrt3}{2}sinx-\frac{1}{2}cosx)=4sin(x-\frac{\pi }{6})$

$\sqrt3sin2x+cos2x=2(\frac{\sqrt3}{2}sin2x+\frac{1}{2}cos2x)=2sin(2x+\frac{\pi}{6})$

Thay vào

$(1)$ ta có

$2sin(x-\frac{\pi}{6})+sin (2x+\frac{\pi}{6})=1$

Đặt

$x-\frac{\pi}{6}=a \rightarrow2x+\frac{\pi}{6}=2a+\frac{\pi}{2}$

$\Rightarrow2sina+sin(2a+\frac{\pi}{2})=1$

$\Rightarrow2sina+cos2a=1$

$\Rightarrow2sin a+1-2sin^2a=1$

Vậy

$sina=0$ hoặc

$sina=1$

Nếu

$sina=0\Rightarrow sin(x-\frac{\pi}{6})=0\Rightarrow x=k\pi+\frac{\pi}{6}$

Nếu

$sina=1\Rightarrow sin (x-\frac{\pi}{6})=1\Rightarrow x=2k\pi+\frac{\pi}{2}+\frac{\pi}{6}=2k\pi+\frac{2\pi}{3}$

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