Question

решить уравнение
sint+cos2t=-1

Answer 1

sint+cos2t=-1
sint+1-2sin²t+1=0
2sin²t-sint-2=0
sint=a
2a²-a-2=0
D=1+16=17
a1=(1-√17)/4⇒sint=(1-√17)/4⇒t=(-1)^n*arcsin[(1-√17)/4]+πn,n∈z
a2=(1+√17)/4⇒sint=(1+√17)/4>1 нет решения

Answer 2

$sint+cos2t=-1 \sint+1-2sin^2t=-1 \-2sin^2t+sint+2=0 \2sin^2t-sint-2=0 \sint=y, y in [-1;1] \2y^2-y-2=0 \D=1+16=17 \y_1= frac{1+sqrt{17}}{4} , sqrt{17}approx 4,1 = textgreater y_1 notin [-1;1] \y_2= frac{1-sqrt{17}}{4} \sint=frac{1-sqrt{17}}{4} \t=(-1)^{n}acrsin(frac{1-sqrt{17}}{4} )+pi n, n in Z$