Using the sum-to-product formulas,
sin6θ + sin2θ = 2 sin4θ cos2θ, so we have
sin6θ + sin4θ + sin2θ = 0
2sin4θ cos2θ + sin4θ = 0
sin4θ (2cos2θ + 1) = 0
sin4θ = 0
cos2θ = -1/2
now finish it off. Post your work if you get stuck.
The number of values of 0 = [0, π] satisfying the equation
sin6theta + sin4theta + sin2theta = 0 is
4 answers
I got n = 0 , 1 which satisfy the given equation
I have no idea what n=0,1 has to do with values for θ
if sin4θ = 0, then
θ = 0+n*2π/4 , π + n*2π/4 = 0, π/4, π/2, 3π/4, π
if cos2θ = -1/2, then
2θ = 2π/3 + n*2π, 4π/3 + n*2π, so
θ = π/3, 2π/3
if sin4θ = 0, then
θ = 0+n*2π/4 , π + n*2π/4 = 0, π/4, π/2, 3π/4, π
if cos2θ = -1/2, then
2θ = 2π/3 + n*2π, 4π/3 + n*2π, so
θ = π/3, 2π/3
I mean total number of values that satisfy given equation is 2. As theta belongs to interval 0,2pi
Am i right or wrong??
Am i right or wrong??