3^2 = 9
21^2 = 441
9 x 441 =
3969
Are
you familiar with that?
------------
amp = 4
and
period=2 pi /4 = pi /2
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
The result of adding the answer together is 1
