3x-2≥0,3x≥2,x≥2/3
x+2>0,x>-2
x≥2/3
x принадлежит [2/3;беск.)
Ответ:
Объяснение:
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2sinxcosx+2cos^2x-2sin^2x-cos^2x-sin^2x=0
2sinx*cosx+cos^2x-3sin^2x=0
-3tg^2x+2tgx+1=0
D=4+4*3*1=16
t=(-2+4)/-6=-1/3
t=(-2-4)/-6=1
tgx=(-1/3) tgx=1
x=-arctg1/3+pik x=pi/4+pin
(8y-5-2y-1-10-y)/7y=(5y-16)/7y
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