Question

Найдите наименьшее значение функции y=2sinx+cos2x на отрезке [0; 5pi/6]

Answer 1

y`=2cosx-2sin2x=0
2cosx-4sinxcosx=0
2cosx(1-2sinx)=0
cosx=0⇒x=π/2+πn,n∈Z
x=π/2∈[0;5π/6]
sinx=1/2⇒x=π/6+2πk U x=5π/6+2πk,k∈Z
x=π/6∈[0;5π/6] U x=5π/6∈[0;5ππ/6]
y(0)=2*0+1=1наим
y(π/6)=2*1/2+1/2=1,5
y(π/2)=2*1-1=1наим
y(5π/6)=2*1/2 +1/2=1,5

Answer 2

Благодарю!

Answer 3

sinx=1/2= (-1)^n*pi/6+pin