find the general solution of the equation `(sqrt3-1)sintheta+(sqrt3 + 1)costheta = 2`

Explanation

Given equation is
`" "(sqrt(3)-1)costheta+(sqrt(3)+1)sintheta=2" "…(i)` ltBrgt Put `" "sqrt(3)-1=rsinalpha and sqrt(3)+1=rcosalpha`
`therefore" "r^(2)=(sqrt(3)-1)^(2)+(sqrt(3)+1)^(2)`
`rArr" "=3+1-2sqrt(3)+3+1+2sqrt(3)`
`rArr" "r^(2)=8`
`therefore" "r=2sqrt(2)`
`now, " "tanalpha=(sqrt(3)-1)/(sqrt(3)+1)=(tan""(pi)/(3)-tan""(pi)/(4))/(1+tan""(pi)/(3)*(pi)/(4))`
` rArr" "tanalpha=tan((pi)/(3)-(pi)/(4))`
`rArr" "tanalpha=tan""(pi)/(12)`
`therefore" " alpha=(pi)/(12)`
From Eq. (i), `rsinalphacostheta+rcosalphasintheta=2` ltBrgt `" "r[sin(theta+alpha)]=2` ltBrgt `rArr" "sin(theta+alpha)=(2)/(2sqrt(2))`
`rArr" "sin(theta+alpha)=(1)/(sqrt(2))`
`rArr" "sin(theta+alpha)=sin""(pi)/(4)theta+alpha=npi+(-1)^(n)(pi)/(4)`
`" "theta=npi+(-1)^(n)*(pi)/(4)-(pi)/(12)`
Alternate method
`" "(sqrt(3)-1)costheta+(sqrt(3)+1)sintheta+2`
Put `" "sqrt(3)-1=rcosalphaand sqrt(3)+1=rsin alpha`
`therefore" "r=2sqrt(2)`
Now, ` " "tanalpha=(sqrt(3)+1)/(sqrt(3)-1)=(1+(1)/(sqrt(3)))/(1-(1)/(sqrt(3)))`
` rArr" "tanalpha=(tan""(pi)/(4)+tan""(pi)/(6))/(1-tan""(pi)/(4)*tan""(pi)/(6))`
`rArr" "tanalpha=tan((pi)/(4)+(pi)/(6))rArrtanalpha=tan""(5pi)/(12)`
`therefore" "alpha=(5pi)/(12)`
From Eq. (i), `rcosalphacostheta+rsinalphasintheta=2`
`r[cos(theta-alpha)]=2`
`rArr" "cos(theta-alpha)=(2)/(2sqrt(2))`
`rArr" "cos(theta-alpha)=(1)/(sqrt(2))`
`rArr" "cos(theta-alpha)=cos""(pi)/(4)`
`rArr" "theta-alpha=2npipm(pi)/(4)`
`therefore" "theta=2npipm(pi)/(4)+(5pi)/(12)`