Question

Sa se determine solutia ecuatiei 2cos^3 x - sin^2 x *cos x - sin^3 x=0

Answer 1

 

$\displaystyle\bf\\2cos^3 x - sin^2 x *cos x - sin^3 x=0~~~\Big|:cos^3x\\\\\frac{2cos^3 x}{cos^3x}-\frac{sin^2 x *cos x}{cos^3x}-\frac{sin^3 x}{cos^3x}=0\\\\2-tg^2x-tg^3x=0\\\\-tg^3x-tg^2x+2=0\\\\tg^3x+tg^2x-2=0\\\\t^3+t^2-2=0\\\\(t^3-1)+(t^2-1)=0\\\\(t-1)(t^2+t+1)+(t-1)(t+1)=0\\\\(t-1)(t^2+t+1+t+t)=0\\\\(t-1)(t^2+2t+2)=0\\\\Doar (t-1)=0 are radacina reala.\\\\t-1=0\\\\t=1\\\\tg\,x=1\\\\\boxed{\bf~x=\frac{pi}{4}+k\pi}$