Question

Решить уравнение 6sin^2x+5cosx-2=0 , принадлежащие отрезку [2П;7П/2]

Answer 1

$6sin^2x+5cos x-2=0\6(1-cos^2x)+5cos x-2=0\6-6cos^2x+5cos x-2=0\-6cos^2x+5cos x+4=0;;;;;times(-1)\6cos^2x-5cos x-4=0\cos x=t,;cos^2x=t^2,;tin[-1;;1]\\6t^2-5t-4=0\D=25-4cdot6cdot(-4)=25+96-121\t_{1,2}=frac{5pm11}{12}\t_1=-frac12\t_2=frac43>1;-;He;nogx.\cos x=-frac12Rightarrow boxed{x=pmfrac{2pi}3+2pi n,;ninmathbb{Z}}\\\xinleft[2pi;;frac{7pi}2right]:\\2pileqpmfrac{2pi}3+2pi nleqfrac{7pi}2$

$begin{array}{ccc}2pileqfrac{2pi}3+2pi nleqfrac{7pi}2&|&2pileq-frac{2pi}3+2pi nleqfrac{7pi}2\&|&\2pi-frac{2pi}3leq2pi nleqfrac{7pi}2-frac{2pi}3&|&2pi+frac{2pi}3leq2pi nleqfrac{7pi}2+frac{2pi}3\&|&\frac{4pi}3leq2pi nleqfrac{17pi}6&|&frac{8pi}3leq2pi nleqfrac{25pi}6\&|&\frac23leq nleqfrac{17}{12}&|&frac43leq nleqfrac{25}{12}\&|&\n=1&|&n=2\&|&\boxed{x_1=frac{2pi}3+2pi=frac{8pi}3}&|&boxed{x_2=-frac{2pi}3+4pi=frac{10pi}3}end{array}$