1) 3tg2x + √3 = 02) 4sin²x - 12cosx + 3 = 03) sin x·cos x - √3 cos² x = 04) sin 3x + sin...

Решите задачу:

$3tg2x+ \sqrt{3}=0\\3tg2x= \sqrt{3} \\ tg2x= \frac{\sqrt{3} }{3} \\ 2x= \frac{ \pi }{6}+ \pi n \\ x= \frac{ \pi }{12}+ \frac{ \pi }{2}n$

$4sin^{2}x-12cosx+3=0\\4-4cos^{2}x-12cosx+3=0\\-4cos^{2}x-12cosx+7=0\\ cosx=t\\-4t^{2}-12t+7=0\\D=(-12)^{2}-4*(-4)*7=144+112=256 \\ x_{1}= \frac{12+16}{2*(-4)}=-3,5 \\ \\x_{2}= \frac{12-16}{2*(-4)}=1 \\ cosx \neq -3,5 \\ cosx=1 \\ x= \pi +2 \pi n$

$sinx*cosx- \sqrt{3}cos^{2}x=0 \\ cosx(sinx- \sqrt{3}cosx)=0\\cosx=0\\x= \frac{ \pi }{2}+ \pi n \\ sinx- \sqrt{3}cosx=0|:cosx\\tgx-\sqrt{3}=0\\tgx=\sqrt{3} \\ x= \frac{ \pi }{3}+ \pi n$

$sin3x+sinx=0\\sin(2x+x)+sinx=0\\sin2xcosx+sinxcos2x+sinx=0\\ sinxcos^{2}x+sinxcos^{2}x-sin^{3}x+sinx=0 \\ 2sinxcos^{2}x-sin^{3}x+sinx=0\\sinx(2cos^{2}x-sin^{2}x+1)=0 \\ sinx=0\\x= \pi n\\2cos^{2}x-sin^{2}x+1=0\\2-2sin^{2}x-sin^{2}x+1=0\\-3sin^{2}x+3=0\\-3sin^{2}x=-3\\3sin^{2}x=3\\sin^{2}x=1\\sinx=+-1\\x_{1}= \frac{ \pi }{2}+2 \pi n$