Trigonometric identity, can you help

Let k_1, k_2 and k_ 3 be real numbers such that, for every real x, k_1 sen(x) + k_2 sen(x + 1) + k_ 3 sen(x + 2) = 0. Does this imply that k_1 = k_2 = k_ 3 = 0? Why or what not?

Thank you

I'm assuming that sen(x) is the sine function.
So you have k1*sen(x) + k2*sen(x+1) + k3*sen(x+2) = 0. Expand this using the identity
sen(a + b) = sen(a)cos(b) + cos(a)sin(b) to get
k1*sen(x) + k2*(sen(x)*cos(1) + cos(x)*sen(1)) + k3*(sen(x)*cos(2) + cos(x)*sen(2)) = 0, therefore
(k1 + k2*cos(1) + k3*cos(2))*sen(x) + (k2*sen(1) + k3*sen(2))*cos(x) = 0.
Now, since sen(x) and cos(x) are orthogonal functions, when we have an equation of the form
A*sen(x) + B*cos(x) = 0 then we must have A=0 and B=0. Therefore we have the following system of equations:

k1 + k2*cos(1) + k3*sen(2) = 0 and k2*sen(1) + k3*sen(2) = 0. This gives an infinite number of possible solution sets for k1, k2, k3, including 0,0,0.
To be more specific, from the second equation we get k3 = -k2 * sin(1)/sin(2) and after putting this into the first equation of our system we get k2 = k1*sin(2) / ((sin(1)*cos(2) - cos(1)*sin(2)). So we can choose any value for k1 and then calculate what k2 and k3 would need to be in order to satisfy the original equation you gave.