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2014-04-14T17:23:31+03:00
Cos2x=1-2sin²x     1-2sin²x=sinx⇔2sin²x+sinx-1=0,faci substitutia sinx=t,t∈[-1,1]           2t²+t-1=0,solutiile vor fi t1=1/2⇒x=π/6 si x=2π/3
t2=-1⇒x=3π/2
Deci solutiile vor fi π/6,2π/3 si 3π/2
2 5 2
2014-04-14T18:27:39+03:00
Cos2x=1-2sin²x    
1-2sin²x=sinx⇔2sin²x+sinx-1=0   ;   faci substitutia sinx=t,t∈[-1,1]          
 2t²+t-1=0,solutiile vor fi t1=1/2⇒x=π/6 si x=2π/3
         --------->  t2=-1⇒x=3π/2
solutiile sunt: π/6,2π/3 si 3π/2