Answer:
a) The general solution θ = nπ±π/6
The all solutions are - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…
b) The solutions in the interval [ 0,2π)
θ = - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6
Step-by-step explanation:
<u><em>Step( i)</em></u>:-
Given an equation 2 cos (2θ)-1 =0
2 cos (2θ) = 1
cos(2θ) = 1/2
cos(2θ) = cos (π/3)
<em>Step(ii):-</em>
<em>a) </em>
The general solution of cos θ = cos ∝ is given by
θ = 2nπ±∝
The general solution of cos(2θ) = cos (π/3) is
2θ = 2nπ±π/3
θ = nπ±π/6
Put n=0 θ = ±π/6
Put n =1 θ = π±π/6
θ = π-π/6 = 5π/6
θ = π+π/6 = 7π/6
put n =2
θ = 2π±π/6
θ = 2π-π/6 = 11π/6
θ = 2π+π/6 = 13π/6
The all solutions are - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…
b)
The solutions in the interval [ 0,2π)
θ = - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6
<u><em>Final answer</em></u>:-
a)
The all solutions are - π/6 ,π/6 ,5π/6 ,7π/6 ,11π/6 ,13π/6 ...…
b)
The solutions in the interval [ 0,2π)
θ = - π/6 ,π/6 , 5π/6 , 7π/6 , 11π/6
