Question #5bc91

Question

969 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-01T22:16:16

#x_1=pi/12#, #x_2=(3pi)/4# and #x_3=(17pi)/12#

#2(cos3x)^2-3sqrt2cos3x+2=0#

#(sqrt2cos3x)^2-3sqrt2cos3x+2=0#

After using #y=sqrt2cos2x# transformation, this equation became,

#y^2-3y+2=0# or #(y-1)*(y-2)=0#

Hence #y_1=1# and #y_2=2#

For #y=2#, #sqrt2*cos3x=2# or #cos3x=sqrt2#. This is impossible.

For #y=1#, #sqrt2*cos3x=1# or #cos3x=sqrt2/2#. Hence #3x=pi/4+2pi*k# or #x=pi/12+(2pi*k)/3#

Thus solutions of it are #x_1=pi/12# (for #k=0#), #x_2=(3pi)/4# (for #k=1#) and #x_3=(17pi)/12# (for #k=2#)