How do you solve # 2 cos^2(4x)-1=0#?

2 Answers
Ratnaker Mehta
Jul 25, 2016

The Soln. Set#={kpi/2+-pi/16} uu {kpi/2+-3pi/16}, k in ZZ#.

Explanation:

We recall that the soln. to the eqn. #costheta=cosalpha# is

#theta=2kpi+-alpha, where, k in ZZ#

Now, the given eqn. is #2cos^2(4x)-1=0 rArr cos4x=+-1/sqrt2#

#:. cos4x=1/sqrt2=cospi/4 rArr 4x=2kpi+-pi/4 rArr x=kpi/2+-pi/16#, &,

#cos4x=-1/sqrt2=cos3pi/4 rArr 4x=2kpi+-3pi/4#

#rArr x=kpi/2+-3pi/16#

Thus, the Soln. Set#={kpi/2+-pi/16} uu {kpi/2+-3pi/16}, k in ZZ#.

Nghi N.
Jul 25, 2016

#pi/16 + (kpi)/4#
#(3pi)/16 + (kpi)/4#

Explanation:

Apply the trig identity:
#cos 2x = 2cos^2 x - 1#
#cos 8x = 2cos^2 (4x) - 1 = 0#
#cos 8x = 0#
Trig table and unit circle give 2 solutions:
a. #8x = pi/2 + 2kpi#
#x = pi/16 + (2kpi)/8 = pi/16 + (kpi)/4#
b. #8x = (3pi)/2 + 2kpi#
#x = (3pi)/16 + (kpi)/4#

Check.
#x = pi/16# --> #4x = pi/4# --> #2cos^2 pi/4 = 2(1/2) - 1 = 0# .OK
#x = (3pi)/16 -> 4x = (3pi)/4 --> 2cos^2 ((3pi)/4) = 2(-sqrt2/2)^2 - 1 = 0# OK

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