How do you solve #3 cos x + 3 = 2 sin ^2 x# over the interval 0 to 2pi?

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Answer 1 · 2016-02-12T07:31:01

#3 cos x + 3 = 2 sin ^2 x#

We know that

#Sin^2 x + cos ^ 2 x= 1#
#=>sin^2 x = 1 - cos^2 x #

Lets substitute this in our equation

#3 cos x + 3 = 2(1 - cos^2x)#

#3cosx + 3 = 2 - 2cos^2 x#

Bring everything over to one side

#2cos^2 x + 3 cos x + 1 = 0#

#Let cos x = a#

#=> 2a^2 + 3a + 1 = 0#

#alpha + beta = -3/2#
#alpha*beta = 1/2#

Solving we get

The roots are

#alpha = -1#

#beta = -1/2#

#therefore cos x = -1 or -1/2#

#=>x = 180 or 120#
#=>x = pi or 2/3 pi#

These both lie in the given interval and hence are correct solutions