Question

f (x)= 10cosx + sin2x - 6 = 0 найти критические точки

Answer 1

f'(x)=(10cosx)'+(sin2x)'-6'=-10sinx+2cos2x
-10sinx+2cos²x-2sin²x=0
-10sinx+2-4sin²x=0
sinx=t
4t²+10t-2=0
2t²+5t-1=0
D=25+8=33
t1=(-5+√33)/4
t2=(-5-√33)/4
1)sinx=t2 ;x€∅
2)sinx=(-5+√33)/4
x=(-1)ⁿarcsin(-5+√33)/4+πn; n€Z