Question

Q-The general solution of inequality cos 2 x≤- sin x is

Answer 1

Answer:

x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I

Step-by-step explanation:

1−2sin2  x≤−sin x    ⇒    (2sin x+1)(sin x−1)≥0

sin x≤−1/2    or    sin x≥1

−5π/6+2nπ≤x≤−π/6+2nπ    or    , n ϵ I x=(4n+1)π/2, n ϵ I⇒    -5π6+2nπ≤x≤-π6+2nπ    or    , n ϵ I x=4n+1π2, n ϵ I     (as sin x = 1 is valid only)

In general⇒    In general    x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I