Question #286cd

1 Answer
Jan 22, 2018

# f(x) = 1/(1 - 2sin^2 x) >= 2#
#1/2 >= (1 - 2sin^2 x)#
#2sin^2 x >= 1/2# --> #sin^2 x >= 1/4#
Note that the term #(1 - 2sin^2 x)# should be positive, in order for f(x) to be > 2. That means
#2sin^2 x < 1#, --> #sin^2 x < 1/2#.
Finally,
#1/4 <= sin^2 x < 1/2# --> take the square root -->
#1/2 <= Isin xI < sqrt2/2#
Solve the above inequality in 2 steps:
a. #1/2 <= sin x < sqrt2/2#
Trig table gives and unit circle -->
#sin (pi/6) = sin ((5pi)/6) = 1/2# and
#sin (pi/4) = sin ((3pi)/4) = sqrt2/2#
Accordingly, on the unit circle, the arc x varies inside half closed intervals:
#[pi/6, pi/4) and ((3pi)/4, (5pi)/6].#
b. # - 1/2 >= sin x > - sqrt2/2#
Trig table and unit circle -->
#sin ((7pi)/6) = sin ((11pi)/6) = - 1/2#, and
#sin ((5pi)/4) = sin ((7pi)/4) = - sqrt2/2#
Accordingly, on the unit circle, the arc x varies inside half closed intervals:
# [(7pi)/6, (5pi)/4)# and #((7pi)/4, (11pi)/6]#
Note. The sign #>=# explains the half close intervals. This means the arc: # pi/6, (5pi)/6, (7pi)/6, and (11pi)/6# are included in the solution set.