Question 1056163
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<pre>
sin(y+&#960;/6)cos(y+&#960;/6)= cos^2(y+&#960;/6)

sin(y+&#960;/6)cos(y+&#960;/6)/cos^2(y+&#960;/6)=1

sin(y+&#960;/6)/cos(y+&#960;/6)=1

tan(y+&#960;/6)=1

tan(y+&#960;/6)=tan(&#960;/4)

(y+&#960;/6) = {{{pi/4}}},  {{{5pi/4}}}  (since tan is periodical with the period {{{pi}}})

y = &#960;/4-&#960;/6,   y = {{{5pi/4 - pi/6}}}   (two solutions)

y = {{{pi/12}}},  y = {{{13pi/12}}}   (two solutions)
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