How do I solve for 0º ≤ x < 360º using the equation cos (x+45º) + cos (x-45º) = √2 ?

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Answer 1 · 2018-02-24T20:03:21

#color(blue)(x=0^@)#

Identities:

#color(red)bb(cos(A+B)=cosAcosB-sinAsinB)#

#color(red)bb(cos(A-B)=cosAcosB+sinAsinB)#

#cos(x+45^@)=cosxcos(45^@)-sinxsin(45^@)#

#cos(45^@)=sqrt(2)/2# #color(white)(888)sin(45^@)=sqrt(2)/2#

#:.#

#cos(x+45^@)=cosxsqrt(2)/2-sinxsqrt(2)/2##color(white)(8888)[1]#

#cos(x-45^@)=cosxcos(45^@)+sinxsin(45^@)#

#cos(x-45^@)=cosxsqrt(2)/2+sinxsqrt(2)/2##color(white)(8888)[2]#

Putting #[ 1 ]# and #[ 2 ]# together:

#cosxsqrt(2)/2-sinxsqrt(2)/2+cosxsqrt(2)/2+sinxsqrt(2)/2=sqrt(2)#

Factor:

#sqrt(2)/2(cosx-sinx+cosx+sinx)=sqrt(2)#

Multiply by #bb2# and divide by #bbsqrt(2)#

#(cosx-sinx+cosx+sinx)=2#

Simplify:

#2cosx=2#

#cosx=1#

#x=arccos(cosx)=arccos(1)=>color(blue)(x=0^@, 360^@)#

For:

#0^@<=x<360^@#

Only #color(white)(8888)color(blue)(0^@)#