Question

Sin³x-3sin²xcosx+2cos³x=0 прооошууу

Answer 1

Sin²x(sinx-cosx)-3cos²x(sinx-cosx)=0
(sinx-cosx)(sin²x-3cos²x=0
sinx-cosx=0/cosx
tgx-1=0
tgx=1⇒x=π/4+πn,n∈z
sin²x-3cos²x=0
(1-cos2x)/2-3(1+cos2x)/2=0
1-cos2x-3-3cos2x=0
-4cos2x=2
cos2x=-1/2
2x=+-2π/3+2πk
x=+-π/3+πk,k∈z

Answer 2

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