How do you solve #2cos^2(2x)+cos(2x)-1=0 #?

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Answer 1 · 2016-03-17T05:12:05

#x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi#,

for any integer #k#.

Let #cos(2x)# be #y#. So the equation becomes a standard quadratic

#2y^2 + y - 1 = 0#

Factorize to find the zeroes.

#(2y - 1) * (y + 1) = 0#

This means that #y = 1/2# or #y = -1#.

Now change back #y# to #cos(2x)#.

So we get

#cos(2x) = 1/2#

or

#cos(2x) = -1#

Next we find the basic angles.

#cos^{-1}(1/2) = pi/3#

#cos^{-1}(-1) = pi#

With that, we solve #cos(2x) = 1/2# first.

#2x = pi/3 + 2kpi or {2pi}/3 + 2kpi#,

for any integer #k#.

#x = pi/6 + kpi or pi/3 + kpi#

Now, we solve #cos(2x) = -1#.

#2x = pi + 2kpi#,

for any integer #k#.

#x = pi/2 + kpi#

Combining the solution gives

#x = pi/6 + kpi or pi/3 + kpi or pi/2 + kpi#,

for any integer #k#.