Question

sinx +radical din3 cosx = -radical 2

Answer 1

impartim prin 2 toata ecuatia
 obtinem o ecuatie echivalenta (adica avand aceleasi solutii)
1/2*sinx+ (√3/2) 8cosx=-√2/2
sinxcosπ/3+sinπ/3 *cosx=-√2/2
 sin (x+π/3°) =-√2/2

x+π/3= kπ+(-1)^k*arcsin(-√2/2)
x+π/3=kπ+(-1)^k*(-π/4)
 x=kπ+(-1)^k*(-π/4)-π/3, unde k∈Z
sau
x=kπ+ (-1)^(k+1)*(π/4)-π/3, k∈Z