Question

Anonim the god 3-4 sin^2 t=0 sin^2-sin t=0 4sin^2 t-1=0 2sin^2 t+sin t=0

Answer 1

$1)3-4 sin^2 t=0 \\4sin^2t=3 \\sin^2t= \frac{3}{4} \\sint=\sqrt{\frac{3}{4} }= \frac{\sqrt{3}}{2} \\t_1= \frac{\pi}{3} +2\pi n,\ n \in Z \\t_2= \frac{2\pi}{3} +2\pi n,\ n \in Z \\sint=-\sqrt{\frac{3}{4} }=- \frac{\sqrt{3}}{2} \\t_3= -\frac{\pi}{3} +2\pi n,\ n \in Z \\t_4= -\frac{2\pi}{3} +2\pi n,\ n \in Z \\2)sin^2t-sin t=0 \\sint(sint-1)=0 \\sint=0 \\t_1=\pi n,\ n \in Z \\sint=1 \\t_2= \frac{\pi}{2} +2\pi n,\ n \in Z$
$3)4sin^2 t-1=0 \\4sin^2t=1 \\sin^2t= \frac{1}{4} \\sint=\sqrt{ \frac{1}{4} }= \frac{1}{2} \\t_1= \frac{\pi}{6} +2\pi n,\ n \in Z \\t_2= \frac{5\pi}{6} +2\pi n,\ n \in Z \\sint=-\sqrt{ \frac{1}{4} }= -\frac{1}{2} \\t_3= -\frac{\pi}{6} +2\pi n,\ n \in Z \\t_4= -\frac{5\pi}{6} +2\pi n,\ n \in Z \\4)2sin^2 t+sin t=0 \\sint(2sint+1)=0 \\sint=0 \\t_1=\pi n,\ n \in Z \\sint= -\frac{1}{2} \\t_2= -\frac{\pi}{6} +2\pi n,\ n \in Z \\t_3= -\frac{5\pi}{6} +2\pi n,\ n \in Z$