Question 1189597

given:

{{{sin(theta)+cos(theta)= 1/sqrt(2)}}}

square both sides:

{{{sin^2 (theta) + 2sin(theta)cos(theta) + cos^2 (theta) = 1/2}}}.......{{{2sin(theta)cos(theta)= sin (2theta) }}}
{{{1 + sin (2theta) = 1/2}}}

{{{sin (2theta) = 1/2-1}}}

{{{sin (2theta) = -1/2 }}}


{{{2theta = 210}}}°  or  {{{2theta = 330}}}°
then
{{{theta= 105}}}° or {{{theta= 165}}}°


but the period of {{{sin (2theta)}}} is {{{180}}}°, so other solutions would be

{{{105+180 = 285}}}° and  {{{165+180 = 345}}}


BUT, since we squared our equation, all answers must be verifies

if {{{theta = 105}}}°

{{{sin(105) + cos(105) = 1/sqrt(2)}}}

if {{{theta = 165}}}°

{{{sin (165 )+ cos (165)}}}  ≠ {{{1/sqrt(2)}}}

if {{{theta = 285}}}°

{{{sin(285)+cos(285) }}} ≠ {{{1/sqrt(2)}}}

if {{{theta= 345}}}°

{{{sin (345)+cos(345)  = 1/sqrt(2)}}}


so {{{theta = 105}}}° in quad II

or

{{{theta = 345}}}°, which is in quad IV


answer: (B) θ must be in Quadrant II or IV