Răspuns:
sin²x+2sin2x-5cos²x=0
sin²x+2*2sinxcosx-5cos²x=0
pui conditia ca x∉{0,π,2π} si imparti ecuatia prin sin²x
[tex]\frac{sin^2x}{sin^2x} +4\frac{sinx*cosx}{sin^2x} -5\frac{cos^2x}{sin^2x}=0[/tex]
1+4ctgx-5tg²x=0
-5tg²x+[tex]4\frac{1}{tgx} +1=0[/tex]
-5tg²*tgx+tgx+4=0║·(-1)
5tg³-tgx-4=0
tgx=y
5y³-y-4=0
4y³+y³-y-4=0
4y³-4)+(y³-y)=0
4(y³-1)+y(y²-1)=0
4(y-1)(y²+y+1)+y(y-1)(y+1)=0
(y-1)(4y²+4y+4+y²+y)=0
(y-1)(5y²+5y+4)=0
y-1=0
y=1
revii la substitutie
tgx=1 x1=[tex]\frac{\pi }{4}[/tex] => =[tex]\frac{\pi }{4} \\x2=\frac{\pi }{4} +\pi =\frac{5\pi }{4}[/tex]=
5y²+5y+4=0
Δ=5²-80=25-80= -55<0
Ecuatia nu are solutii reale
Deci x∈{[tex]\frac{\pi }{4};\frac{5\pi }{4} }[/tex]}
Explicație pas cu pas:
