Решите уравнение: 2cos^2x+5sinx+1=0 2 sin^2x/2+19sinx/2-10=0

2cos^2x=cos2x+1cos2x+1+5sinx+1=01-2sin^2x+1+5sinx+1=02sin^2x-5sinx-3=0sinx=t2t^2-5t-3=0D=25+4*2*3=49t1=-2/4=-1/2t2=(5+7) /2=6 (ne mojet) sinx=-1/2x=(-1) ^k+1pi/6+pik sin^2x/2+19sinx/2-10=0[2cos^2x=cos2x+1\\ cos2x+1+5sinx+1=0\\1-2sin^2x+1+5sinx+1=0\\2sin^2x-5sinx-3=0sinx=t\\2t^2-5t-3=0\\D=25+4*2*3=49\\ t1=-2/4=-1/2\\ t2=(5+7) /2=6 (ne mojet) \\sinx=-1/2\\ x=(-1) ^k\pi/6+\pi k\\ 2sin^2x/2+19sinx/2-10=0 \\sinx/2=t\\2t^2+19t-10=0\\D=21^2\\t_1=(-19-21) /4=-10t_2=(-19+21) /4=1/2sinx/2=1/2x/2=(-1) ^k\pi/6+\pi*kx=(-1) ^k\pi/3+2\pi k[/tex]