Question

Va rog explicatimi aceste exerciti cum se fac <>_<> sin($\frac{π}{2}$ -x)+sin($\frac{π}{2}$ +x)+2cos(π-x) =0 si sin²($990^{0}$-x)+2cos(x-$360^{0}$ )=3 ce formule si cum se rezolva acest exercitiu plz raman dator

Answer 1

Ecuatia 1:     
$sin( \frac{\pi}{2}-x)+sin( \frac{\pi}{2}+x)+2cos(\pi-x)=0 \\ \text{aducem la primul cadran: } \\ sin( \frac{\pi}{2}-x) = cos\,x \\ sin( \frac{\pi}{2}+x)=cos\,x \\ cos(\pi-x)=-cos\,x \\ =\ \textgreater \ \;\;\;cos\,x+cos\,x+2(-cos\,x)=0 \\ 2cos\,x-2cos\,x=0 \\ 0=0 \\ =\ \textgreater \ \text{Egalitatea nu depinde de x} \\ =\ \textgreater \ \;\text{x \;poate lua orice valoare}$


Ecuatia 2:
$sin^2(990^o-x)+2cos(x-360^o)=3 \\ \text{Aducem la primul cadran:} \\ sin(990^o-x)=sin(360^o+ 360^o+270^o-x)= \\ =sin(360^o+270^o-x) =sin(270^o-x)=-cos\,x \\ cos(x-360^o)=cos\,x \\ \\ (-cos\,x)^2 + 2cos\,x=3 \\ cos^2x+2cos\,x-3=0 \\ Substitutie:\;\;\;\;z=cos\,x \\ z^2+2z-3=0 \\ z_{12}= \frac{-2 \;\pm\; \sqrt{4+12} }{2} = \frac{-2 \;\pm\; \sqrt{16}}{2} = \frac{-2 \;\pm\; 4}{2} =-1 \;\pm\; 2 \\ z_1 = -1+2=1 \;\;\;=\ \textgreater \ \;\;\;cos\,x=1\;\;\;=\ \textgreater \ \;\;\;x=0+2k\pi \\ z_2=-1-2=-3\;\;\; cos\,x=-3 \;\;\; (imposibil)$