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

Решите задачу:

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