Question

sa se rezolve urmatoarele ecuatii trigonumetrice

Answer 1

$8) \frac{sinx}{cosx}+3 \frac{cosx}{sinx}=4 \\ \frac{sin^2x+3cos^2x-4sinxcosx}{sinxcos}=0 \\ sin^2x+3cos^2x-4sinxcosx=0 si x \neq n \pi ;x \pi \neq \frac{ \pi }{2}+k \pi \\ tg^2x-4tgx+3=0 \\ fie :tgx=a \\ a^2-4a+3=0 \\ a=1;a=3 \\ tgx=1;x= \frac{ \pi }{4}+n \pi \\ tgx=3;x=arctg3+k \pi \\ 10)3(cos^2x-sin^2x)+2cosx=5 \\ 3(2cos^2x-1)+2cosx=5 \\ 6cos^2x+2cosx-8=0 \\ 3cos^2x+cosx-4=0; \\ cosx=--8/6; \\ cosx=1;x=2n \pi$