Question

În mulţimea [0,2π] rezolvaţi ecuaţia sin ^2x -cos^2x =cos x

Answer 1

cosx=-cos2x
cosx+cos2x=0
 2 cos(3x/2)*cos(x/2)=0

3x/2=π/2; 3π/2;5π/2;7π/2...
3x=π;3π;5π;7π...
x={π/3;π;5π/3}

x/2=π/2;3π/2;5π/2;7π/2...
x=π

S in [0;2π]={π/3;π;5π/3}