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Question

giải pt:
1/ $sin^{2}2x$ $cos6x$ + $sin^{2}3x$ = $\frac{1}{2}$ $sin2xsin8x$
2/ $(\sqrt{3} \sin 2x$ + 3)cos x$ = (2cosx - $$\sqrt{3}$$ sinx)$cos2x

score 1 · Answer 1

$(\sqrt{3} \sin 2x + 3)\cos x= (2\cos x - \sqrt{3}\sin x)\cos 2x$

$\Leftrightarrow \sqrt 3( \sin 2x \cos x+ \sin x \cos 2x) +3\cos x -2\cos 2x \cos x=0$

$\Leftrightarrow \sqrt 3 \sin 3x +3\cos x -\cos 3x -\cos x=0$

$\Leftrightarrow \sqrt 3 \sin 3x -\cos 3x =-2\cos x$ chia 2 vế cho $2$ là xong

score 1 · Answer 2

PT 1 $\Leftrightarrow (1-\cos 4x)\cos 6x +1-\cos 6x =\sin 2x \sin 8x$

$\Leftrightarrow 1-\cos 4x \cos 6x -\sin 2x \sin 8x =0$

$\Leftrightarrow 2-\cos 10x -\cos 2x +\cos 10x -\cos 6x=0$

$\Leftrightarrow \cos 6x +\cos 2x -2=0$

$\Leftrightarrow 4\cos^3 2x -3\cos 2x +\cos 2x -2=0$

$\Leftrightarrow 4\cos^3 2x -2\cos 2x -2=0$

$\Leftrightarrow \cos 2x = 1 \Rightarrow x=k\pi;\ k\in Z$

2 Đáp án