toán 11

L

lan_phuong_000

Đề của bạn là:

$sinx + \sqrt{3}.cosx = \sqrt{2 + cos2x + \sqrt{3sinx}}$
ĐK: $sinx + \sqrt{3}.cosx$ \geq 0 \Leftrightarrow x \geq $\dfrac{-\pi}{3}$
\Leftrightarrow $sinx + \sqrt{3}.cosx = \sqrt{3 - 2sin^2x + \sqrt{3sinx}}$
\Rightarrow $sin^2x + 3cos^2x + \sqrt{3}.sin2x = -2sin^2x + 3 + \sqrt{3sinx}$
\Leftrightarrow $sin2x=\sqrt{sinx}$
\Leftrightarrow $x = k\pi$ (tm) (k thuộc N)
 
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