Answer:
Three consecutive multiples are 110,121 and 132 which has the sum of 363. 33C = 363; C = 33, By substituting C = 33 in numbers 11C, 11(C – 1) and 11(C + 1) we get values 110, 121, 132.
Hello :
cos(3<span>θ) = -1
</span>cos(3θ) = cos(<span>π)
</span>general solution in : R is
3θ = (2k+1)π k in Z
θ = (2k+1)π/3
if θ in <span> [0 , 2</span>π] the solutions are :
k=0 : x = π/3
k=1 : x = π
k=2 : x = 5π/3
