Question

40 puncte , exercititul este in poza atasata.

Answer 1

Răspuns:

sin²x+2sin2x-5cos²x=0

sin²x+2*2sinxcosx-5cos²x=0

pui conditia ca x∉{0,π,2π} si imparti ecuatia prin sin²x

$\frac{sin^2x}{sin^2x} +4\frac{sinx*cosx}{sin^2x} -5\frac{cos^2x}{sin^2x}=0$

1+4ctgx-5tg²x=0

-5tg²x+$4\frac{1}{tgx} +1=0$

-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=$\frac{\pi }{4}$  => =$\frac{\pi }{4} \\x2=\frac{\pi }{4} +\pi =\frac{5\pi }{4}$=

5y²+5y+4=0

Δ=5²-80=25-80= -55<0

Ecuatia  nu   are    solutii reale

Deci x∈{$\frac{\pi }{4};\frac{5\pi }{4} }$}

Explicație pas cu pas:

Answer 2

Răspuns:

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Explicație pas cu pas: