Question #3aca5

1 Answer
Jun 16, 2017

#(7pi)/6; (11pi)/6#

Explanation:

1/(sec^2 x) + 3sin (x/2)cos (x/2) = 0
since #1/(sec^2 x) = cos^2 x = 1 - sin^2 x#, and
#3sin (x/2)cos (x/2) = (3/2)sin x#, therefore:
#1 - sin^2 x + (3/2)sin x = 0#
#- 2sin^2 x + 3sin x + 2 = 0#
Solve this quadratic equation for sin x.
#D = d^2 = b^2 - 4ac = 9 + 16 = 25# --> #d = +- 5#
There are 2 real roots:
#sin x = -b/(2a) +- d/(2a) = -3/-4 +- 5/4 = 3/4 +- 5/4#
#sin x = 8/4 = 2# (rejected as > 1), and
#sin x = - 2/4 = - 1/2#
Trig table and unit circle give:
#x = (-pi)/6# or #x = (11pi)/6# (co-terminal), and
#x = pi - (- pi/6) = x + pi/6 = (7pi)/6#
Check by calculator.
#x = (7pi)/6 = 210^@# --> cos x = -0.866 --> #cos^2 x = 0.75# -->
x/2 = 105 --> cos (x/2) = -0.258 --> sin (x/2) = 0.966.
cos^2 x + 3sin (x/2).cos (x/2) = 0.75 + 3(-0.258)(0.966) =
= 0.75 - 0.75 = 0. Correct.