Question 1206408


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

Find the solutions in the interval [{{{0}}}, {{{2pi}}}).



{{{1 -2 sin(theta) = cos(2theta)}}}....use identity {{{cos(2theta)=cos^2(theta) - sin^2(theta)}}}

{{{1 -2 sin(theta) = cos^2(theta) - sin^2(theta)}}}

{{{sin^2(theta)-cos^2(theta) -2 sin(theta) +1= 0}}}...use identity {{{cos^2(theta)=1-sin^2(theta)}}}

{{{sin^2(theta)-(1-sin^2(theta)) -2 sin(theta) +1= 0}}}

{{{sin^2(theta)-1+sin^2(theta) -2 sin(theta) +1= 0}}}

{{{2sin^2(theta) -2 sin(theta) = 0}}}...factor

{{{2sin(theta)(sin(theta) -1)  = 0}}}


solutions:

if {{{2sin(theta) = 0}}} =>{{{sin(theta) = 0}}}
if {{{2sin(theta)-1 = 0}}} =>{{{sin(theta) = 1/2}}}


find {{{theta}}}

{{{theta=sin^-1(0)}}}=> {{{theta=0}}} (result in radians) or {{{theta=0}}}°
{{{theta=sin^-1(1/2)}}}=>{{{theta= pi/6}}} (result in radians) or{{{theta=30}}}°

the solutions in the interval [{{{0}}}, {{{2pi}}})
{{{theta=0}}}or {{{theta=0}}}°
{{{theta=pi/6}}} or{{{theta=30}}}°
{{{theta=5pi/6}}} or{{{theta=150}}}°
{{{theta=pi}}} or{{{theta=180}}}°