Question #d89db

1 Answer
Nghi N
Sep 8, 2017

#pi/6; (5pi)/6#

Explanation:

#f(x) = 4cos^2 x - 4sin x - 1 = 0#
Replace #cos^2 x# by #(1 - sin^2 x)#
#4(1 - sin^2 x) - 4sin x - 1 = 0#
#4 - 4sin^2 x - 4sin x - 1 = 0#
Change side. Solve the quadratic equation by the improved quadratic formula in graphic form (Socratic, Google Search):
#4sin^2 x + 4sin x - 3 = 0#
#D = d^2 = b^2 - 4ac = 16 + 48 = 64# --> #d = +- 8#

There are 2 real roots:
#sin x = - b/(2a) +- d/(2a) = - 1/2 +- 1#
#sin x = - 1/2 + 1 = 1/2# and #sin x = - 3/2# (rejected as < - 1)
#sin x = 1/2# --> trig table and unit circle give -->
#x = pi/6#, and #x = pi - pi/6 = (5pi)/6#

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