How do you solve #2\cos^{2} x + \cos 2x + \sin x = 0#?

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Answer 1 · 2016-11-02T02:40:36

We will need the following identities to solve this equation.

#•cos^2x = 1 - sin^2x#
#•cos2x = 1 - 2sin^2x#

Our goal here is to change everything into sine. Here it goes:

#2(1 - sin^2x) + 1 - 2sin^2x + sinx = 0#

#2 - 2sin^2x + 1 - 2sin^2x + sinx = 0#

#0 = 4sin^2x - sinx - 3#

Let #t = sinx#.

#0 = 4t^2 - t - 3#

#0 = 4t^2 - 4t + 3t - 3#

#0 = 4t(t - 1) + 3(t- 1)#

#0 = (4t + 3)(t - 1)#

#t = -3/4 and 1#

#sinx = -3/4 and sinx = 1#

#x = pi + arcsin(3/4), 2pi - arcsin(3/4), pi/2#

Hopefully this helps!