How do you solve `Cos([θ/3]-[pi/4])=1/2`?

`"please remember that " cos (a-b)=cos a*cos b+sin a* sin b`

`cos (theta/3-pi/4)=cos theta*cos pi/4+sin theta*sin pi/4`

`cos pi/4=sin pi/4=sqrt2/2`

`cos((theta/3)-pi/4)=cos (theta/3)*sqrt2/2+sin( theta/3)*sqrt2/2`

`cos(theta/3)*sqrt2/2+sin (theta/3)*sqrt2/2=1/2`

`sqrt2/2(cos (theta/3) +sin (theta/3))=1/2`

`cos (theta/3)+sin (theta/3)=1/cancel(2)*cancel(2)/sqrt2`

`cos ( theta/3) +sin (theta/3)=1/sqrt2" ; "1/sqrt2=sqrt2/2`

`cos (theta/3)+sin (theta/3)=sqrt2/2`

`[cos (theta/3)+sin (theta/3)]^2=[sqrt2/2]^2`

`cos^2(theta/3)+2*sin(theta/3)*cos(theta/3)+cos^2(theta/3)=1/2`

`"so ;" cos^2 a+sin^2 a=1;`

`cos^2(theta/3)+sin^2(theta/3)=1`

`1+2*sin(theta/3)*cos(theta/3)=1/2`

`2*sin(theta/3)*cos(theta/3)=1/2-1`

`2*sin(theta/3)*cos(theta/3)=-1/2`

`"so ; "2*sin a*cos a=sin(2*a)`

`2*sin(theta/3)*cos(theta/3)=sin((2*theta)/3)`

`sin((2theta)/3)=-1/2`

`(2theta)/3=-30^o`

`2theta=-90`

`theta=-45 ^o`

The principal value of `(theta/3-pi/4)`

`=` the principal value of `cos^(-1)(1/2)=pi/3`.

So, this `theta` is given by`(theta/3-pi/4) =(pi/3`. And so, `theta=(7pi)/4`.

The general value is given by

`theta/3-pi/4=2npi+pi/3, n=0,+-1, +-2, =-3, ...`

S0, `theta = (6n+7/4)pi, n=0, +-1, +-2, +-3,...`