Question

Anonim know everything xD
2cos^2 t+sin t+1=0
sin^2 t+3cos t-3=0

Answer 1

$1)2cos^2 t+sin t+1=0 \2(1-sin^2t)+sint+1=0 \sint=y, y in [-1;1] \2-2y^2+y+1=0 \-2y^2+y+3=0 \2y^2-y-3=0 \D=1+24=25=5^2 \y_1= frac{1+5}{4} notin [-1;1] \y_2= frac{1-5}{4} =-1 in [-1;1] \sint=-1 \t=- frac{pi}{2} +2pi n, n in Z \2)sin^2 t+3cos t-3=0 \1-cos^2t+3cost-3=0 \cost=y, y in [-1;1] \-y^2+3y-2=0 \y^2-3y+2=0 \D=9-8=1 \y_1= frac{3+1}{2} =2notin [-1;1] \y_2= frac{3-1}{2} =1in [-1;1] \cost=1 \t=2pi n, n in Z$
$a)sin( frac{pi}{2} +t)-cos(pi+t)=1 \cost-(-cost)=1 \2cost=1 \cost= frac{1}{2} \t_{1,2}=pm frac{pi}{3} +2pi n, n in Z \b)sin(pi+t)+sin(2pi-t)-cos( frac{3pi}{2} +t)+1,5=0 \-sint+sin(-t)-sint+1,5=0 \-3sint=-1,5 \sint= frac{1,5}{3} = 0,5= frac{1}{2} \t_1= frac{pi}{6} +2pi n, n in Z \t_2=frac{5pi}{6} +2pi n, n in Z$
$c)cos( frac{pi}{2} -t)-sin(pi+t)=sqrt{2} \sint-(-sint)=sqrt{2} \2sint=sqrt{2} \sint= frac{sqrt{2}}{2} \t_1= frac{pi}{4} +2pi n, n in Z \t_2= frac{3pi}{4} +2pi n, n in Z \d) sin(pi+t)+cos( frac{pi}{2} +t)=sqrt{3} \-sint-sint=sqrt{3} \sint=- frac{sqrt{3}}{2} \t_1= -frac{pi}{3} +2pi n, n in Z \t_2= -frac{2pi}{3} +2pi n, n in Z$

Answer 2

https://znanija.com/task/27123041 final