Question

3cos2x+2cosx=5
Va rog ajutatima e urgent

Answer 1

3(2cos²x-1)+2cosx=5

6cos²x-3+2cosx-5=0

notam cosx=t

6t²+2t-8=0

3t²+t-4=o

t1=1, cosx=1, x=0

t2=-4/3 nu apartine [-1,1], ecuatia  cos x=-4/3 nu are solutii

Answer 2

   
[tex]\texttt{Formule folosite: } \\ cos\,2x = cos^2x - sin^2x\\ sin^2x+cos^2x = 1~~\Longrightarrow~~ sin^2 = 1-cos^2 \\ \\ \texttt{Rezolvare:} \\ 3cos\,2x + 2cos\,x = 5 \\ 3(cos^2x - sin^2x)+ 2cos\,x = 5 \\ 3(cos^2x - (1-cos^2x))+ 2cos\,x = 5 \\ 3(cos^2x - 1+cos^2x)+ 2cos\,x = 5 \\ 3(2cos^2x - 1)+ 2cos\,x - 5 =0\\ 6cos^2x - 3+ 2cos\,x - 5 =0 \\ 6cos^2x + 2cos\,x - 8 =0 ~~~~~|:2 \\ 3cos^2x + cos\,x - 4 =0 [/tex]


[tex]\displaystyle \\ \texttt{Rezolvam ecuatia de gradul 2} \\ Notam: ~~\boxed{t = cos\;x} \\ 3t^2 + t - 4 =0 \\ \\ t_{12} = \frac{-1 \pm \sqrt{1 - 4\times 3 \times(-4)} }{2\times 3}= \\ \\ = \frac{-1 \pm \sqrt{1+48}}{6}= \frac{-1 \pm \sqrt{49}}{6} = \frac{-1\pm7}{6}\\ \\ t_1 = \frac{-1 + 7}{6} = \frac{6}{6} = \boxed{1} \\ \\ t_2 = \frac{-1 - 7}{6} = \frac{-8}{6} = \boxed{\frac{-4}{3}} ~~\texttt{Solutie eliminata deoarece }t \ \textless \ -1 [/tex]


[tex]\\ \\ \text{t2 se elimina deoarece cosinusul nu poate fi mai mic decat minus unu } \\ \\ t1 = 1\\\Longrightarrow ~~cos\;x=1 \\ \Longrightarrow ~~ \boxed{x = 0 + 2k\pi, ~~k \in N } [/tex]