Question

(sin2 + 2 − √3 sin 2 − cos2 )√(4 − ) = 0
$(sin^2x + 2 - \sqrt{3} sin 2x - cos^2x )\sqrt{x(4 - x )} = 0$

Answer 1

$(sin^2x+2-\sqrt{3}sin2x-cos^2x)\sqrt{x(4-x)} =0\\\\ sin^2x+2-\sqrt{3}sin2x-cos^2x=0\\sin^2x+2(sin^2x+cos^2x)-2\sqrt{3}sinxcosx-cos^2x=0\\sin^2x+2sin^2x+2cos^2x-2\sqrt{3}sinxcosx-cos^2x=0\\3sin ^2x-2\sqrt{3}sinxcosx+cos^2x=0 |\div cos^2x \\3tg^2x-2\sqrt{3}tgx+1=0\\(\sqrt{3}tgx-1)^2=0\\\sqrt{3}tgx-1=0\\tgx=\frac{1}{\sqrt{3}} \\x=arctg \frac{1}{\sqrt{3}}+\pi n=\frac{\pi }{6}+\pi n ;n\in Z \\\\n=-1\\x=-\frac{5\pi}{6} \notin[0;4]\\\\n=0;\\x=\frac{\pi}{6} \in [0;4]\\\\\\n=1;\\x= \frac{7\pi}{6} \in [0;4]$

$n=2\\x=\frac{13\pi}{6} \notin [0;4]$

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$2)\sqrt{x(4-x)}=0\\x(4-x)=0\\x_1=0 ; x_2=4\\\\Proverka:\\x=0\\0(4-0)=0\\0=0\\\\x=4\\4(4-4)=0\\4*0=0\\0=0$

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Ответ:

           $x=0 \\x=4\\x=\frac{\pi}{6}\\x=\frac{7\pi}{6}$