Question

5sin^2x+cosx-1=0 ///////

Answer 1

$5sin^{2}x+cosx-1=0 \\ 5(1-cos^{2}x)+cosx-1=0 \\ -5cos^{2}x+cosx+4=0 \\ cosx=u \\ -5u^{2}+u+4=0 \\ D:1+80=81 \\ u_1,_2= \frac{1\pm 9}{10} \\ u_1=1 \\ u_2=- \frac{4}{5} \\ \\ 1)cosx=1 \\ x=2\pi n,n\in Z;$

$2)cosx= -\frac{4}{5} \\ x=\pm arccos(- \frac{4}{5} )+2\pi n \\ x=\pm(\pi-arccos \frac{4}{5})+2\pi n,n\inZ$