Let θ∈0,4π satisfy the equation sinθ+2sinθ+3sinθ+4=6 .If the sum of all the values of θ is of the form kπ , then the value of k is
6
5
4
2
We have sinθ+2sinθ+3sinθ+4=6
Since -1≤sinθ≤1, LHS≥6
But RHS=6⇒sinθ=-1
Since θ∈0,4π⇒θ=3π2,7π2
∴sum of values of θ=3π2+7π2=5π=kπ ⇒k=5