См. вложение
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2cos^3x=cos(-x)
2cos^3x=cosx
2cos^3x-cosx=0
cosx(2 cos^2x-1)=0
<span>
cosx=0
</span><span>x= π/2 + 2πk </span><span>
и</span>
2
cos^2x-1=0
2 cosx=1
<span>
сosx=1/2</span>
<span>x=π/6+2πk
</span>x=5π/6+2πk
<span>Ответ<span>: π/2
+ 2πk ; π/6+2πk ; 5π/6+2πk</span></span>
А) х²+4х=0
х(х+4)=0
х=0 и х+4=0
Х=-4
Б)-х²-8х=0
-х(х-8)=0
-х=0 и х-8=0
х=0 и х=8
в)15х-х²=0
х(15-х)=0
х=0 и 15-х=0
х=15