How do you solve #3cos^2 x - 5cos x -1= 3 sin^2 x#?

1 Answer
Dilini Peiris · Becca M.
Sep 1, 2015

#x = 2npi +- 2pi/3#

The description looks long, but I've written every single step so that anyone can understand!

Explanation:

#3cos^2x−5cosx−1=3sin^2x#

#3(cos^2x - sin^2x) - 5cosx-1=0#

#3(cos^2x-1+cos^2x) - 5cosx -1=0#

#6cos^2x -5cosx -4=0#

from here its just a quadratic equation. you divide it to its factors.

#(3cosx-4) (2cosx +1) =0#

#3cosx-4=0# or #2cosx+1=0#
from here u get two possibilities

#cos x = 4/3# or #cos x = -1/2#
the first answer cannot be true because the cosines should be between -1 and +1
if u remember the cosine curve u should be able to understand what im talking about.

so we are left with the latter answer

#cosx= -pi/3#
#cos x = (pi - pi/3)# you dont really have to show this step.. u can do it in your mind ;)
#cos x =2pi/3#

according to the trig equations #theta = 2npi +- theta#
just replace the correct places of this equations with our answer
so the answer is
#x = 2npi +- 2pi/3#

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