Question

For anоnim 2sin^2 t+3sin-2=0 2cos^2 t - 5cos t +2=0 2sin^2 t+sin t-1=0 4cos^2 t+9 cos t+5=0

Answer 1

$1)2sin^2 t+3sint-2=0 \\sint=y,\ y \in [-1;1] \\2y^2+3y-2=0 \\D=9+16=25=5^2 \\y_1= \frac{-3+5}{4} = \frac{1}{2} \in [-1;1] \\y_2= \frac{-3-5}{4} \notin [-1;1] \\sint= \frac{1}{2} \\t_1= \frac{\pi}{6} +2\pi n,\ n \in Z \\t_2= \frac{5\pi}{6} +2\pi n,\ n \in Z$
$2)2cos^2 t - 5cos t +2=0 \\cost=y,\ y \in [-1;1] \\2y^2-5y+2=0 \\D=25-16=9=3^2 \\y_1= \frac{5+3}{4} =2 \notin [-1;1] \\y_2= \frac{5-3}{4} = \frac{1}{2} \in [-1;1] \\cost= \frac{1}{2} \\t_{1,2}=\pm \frac{\pi}{3} +2\pi n,\ n \in Z$
$3)2sin^2 t+sin t-1=0 \\sint=y,\ y \in [-1;1] \\2y^2+y-1=0 \\D=1+8=9=3^2 \\y_1= \frac{-1+3}{4} = \frac{1}{2} \in [-1;1] \\y_2= \frac{-1-3}{4} =-1\in [-1;1] \\sint= \frac{1}{2} \\t_{1}= \frac{\pi}{6} +2\pi n,\ n \in Z \\t_2= \frac{5\pi}{6} +2\pi n,\ n \in Z \\sint=-1 \\t_3= -\frac{\pi}{2} +2\pi n,\ n \in Z$
$4)4cos^2 t+9 cos t+5=0 \\cost=y,y \in [-1;1] \\4y^2+9y+5=0 \\D=81-80=1 \\y_1= \frac{-9+1}{8} =-1 \in [-1;1] \\y_2= \frac{-9-1}{8} \notin [-1;1] \\cost=-1 \\t=\pi +2\pi n,\ n \in Z$