Question #cf5ae

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Answer 1 · 2017-03-10T14:38:06

I tried this:

Ok, it should be:
#2cos^2(x)-1=0#
rearrange:
#cos^2(x)=1/2#
#cos(x)=+-1/sqrt(2)=+-sqrt(2)/2#
(where at the end I rationalized).

Now we need to find suitable values for #x# so that the cosine will be equal to #+-sqrt(2)/2#:
Have a look:
enter image source here

At least in the interval #[0,2pi]# we will get 4 possibilities:
A) #x=pi/4#
B) #x=3/4pi#
C) #x=5/4pi#
D) #x=7/4pi#

this will be repeated periodically every #2pi#

Answer 2 · 2017-03-10T14:47:30

#" The Soln. Set="{2kpi+-pi/4 : k in ZZ}uu{2kpi+-3pi/4 : k in ZZ}.#

#2cos^2x-1=0#

#rArr 2cos^2x=1, or, cos^2x=1/2.#

#rArr cosx=+-1/sqrt2.#

Knowing that, #costheta=cosalpha rArr theta=2kpi+-alpha, k in ZZ.#

# cosx=1/sqrt2=cos(pi/4) rArr x=2kpi+-pi/4, k in ZZ.#

# cosx=-1/sqrt2=cos(3pi/4) rArr x=2kpi+-3pi/4, k in ZZ.#

#:." The Soln. Set="{2kpi+-pi/4 : k in ZZ}uu{2kpi+-3pi/4 : k in ZZ}.#

Enjoy Maths.!