Solve the equation #(sqrt3-1)sintheta+(sqrt3+1)costheta=2#?

2 Answers
Jun 14, 2017

#theta=npi+-(-1)^npi/4-(5pi)/12#, where #n# is an integer.

Explanation:

To solve the equation #(sqrt3-1)sintheta+(sqrt3+1)costheta=2#

let us divide each term by #sqrt((sqrt3-1)^2+(sqrt3+1)^2)#

i.e. #sqrt(2((sqrt3)^2+1^2))=sqrt8=2sqrt2# and we get

#(sqrt3-1)/(2sqrt2)sintheta+(sqrt3+1)/(2sqrt2)costheta=2/(2sqrt2)=1/sqrt2#

Let #(sqrt3-1)/(2sqrt2)=cosalpha#, then #(sqrt3+1)/(2sqrt2)=sinalpha#

(as their squares add up to #1# - see the way we divided each term by #2sqrt2#)

and then #tanalpha=(sqrt3+1)/(sqrt3-1)=tan((5pi)/12)#

and our equation becomes

#sinthetacos((5pi)/12)+costhetacos((5pi)/12)=sin(pi/4)#

or #sin(theta+(5pi)/12)=sin(pi/4)#

Hence #theta+(5pi)/12=npi+-(-1)^npi/4#, where #n# is an integer.

or #theta=npi+-(-1)^npi/4-(5pi)/12#, where #n# is an integer.

#sin(pi/12)=sin(pi/3-pi/4)#

#=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)#
#=sqrt3/2xx1/sqrt2-1/2xx1/sqrt2#

#=(sqrt3-1)/(2sqrt2)#

Again

#cos(pi/12)=cos(pi/3-pi/4)#

#=cos(pi/3)cos(pi/4)+sin(pi/3)sin(pi/4)#
#=1/2xx1/sqrt2+sqrt3/2xx1/sqrt2#

#=(sqrt3+1)/(2sqrt2)#

Now the given equation is

#(sqrt3-1)sintheta+(sqrt3+1)costheta=2#

Dividing both sides by #2sqrt2# we get

#(sqrt3-1)/(2sqrt2)sintheta+(sqrt3+1)/(2sqrt2)costheta=2/(2sqrt2)#

#=>sin(pi/12)sintheta+cos(pi/12)costheta=1/sqrt2#

#=>cos(theta-pi/12)=cos(pi/4)#

#=>theta-pi/12=2npipmpi/4" where "n in ZZ#

So
#=>theta=2npi+pi/4+pi/12" where "n in ZZ#

#=>theta=2npi+pi/3" where "n in ZZ#

Again

#theta=2npi-pi/4+pi/12" where "n in ZZ#

#=>theta=2npi-pi/6" where "n in ZZ#