Question 594183
{{{sqrt(3)cos(x)+sin(x)}}} = {{{1}}}
<pre>
Use the identity

{{{A*cos(theta) + B*sin(theta)}}} = {{{sqrt(A^2+B^2)*sin(theta+alpha)}}}, where {{{tan(alpha)=A/B}}} and {{{A > 0}}}, {{{B > 0}}}

 {{{A=sqrt(3)}}}, {{{B = 1}}}, {{{sqrt(A^2+B^2)=sqrt((sqrt(3))^2 + 1^2) = sqrt(3+1) = sqrt(4) = 2}}}

{{{tan(alpha)=A/B}}}
{{{tan(alpha)=sqrt(3)/1}}}
{{{alpha="60°"=pi/3}}}

{{{sqrt(3)cos(x)+sin(x)}}} = {{{1}}}
{{{sqrt(A^2+B^2)*sin(theta+alpha)}}} = {{{1}}} 
{{{2sin(theta+pi/3)}}} = {{{1}}}
{{{sin(theta+pi/3)}}} = {{{1/2}}}

{{{theta+pi/3}}} = {{{pi/6}}}, {{{5pi/6}}}, {{{13pi/6}}}, {{{17pi/6}}}

{{{theta}}} = {{{pi/6-pi/3}}}, {{{5pi/6-pi/3}}}, {{{13pi/3-pi/3}}}, {{{17pi/6-pi/3}}}

{{{theta}}} = {{{pi/6-2pi/6}}}, {{{5pi/6-2pi/6}}}, {{{13pi/6-2pi/6}}}, {{{17pi/6-2pi/6}}}

{{{theta}}} = {{{cross(-pi/6)}}}, {{{3pi/6}}}, {{{11pi/6}}}, {{{15pi/6}}}

{{{theta}}} =   {{{pi/2}}}, {{{11pi/6}}}, {{{cross(5pi/2)}}}

{{{theta}}} =   {{{pi/2}}}, {{{11pi/6}}}

Edwin</pre>