Question

3sin^2x-cosx+1=0
Решите тригонометрию

Answer 1

$3(1-cos^2(x))-cosx+1=0 \ 3-3cos^2(x)-cosx+1=0 \ -3cos^2(x)-cosx+4=0 \ cosx = t ; t in [-1;1] \ -3t^2-t+4=0 \ D=1+3*4*4=1+48=49 \ t_{1,2} = frac{1б7}{-6} ; t_1 = frac{8}{-6} < -1 ; t_1 neq frac{8}{-6} \ t_2 = frac{-6}{-6} = 1; \ cosx = 1; x = 2pk$