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Question

$text The general solution for end text theta text if end text sin invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses plus cos invisible function application open parentheses theta plus fraction numerator 5 pi over denominator 6 end fraction close parentheses equals 2 blank$ is

$2 n π + \frac{7 π}{6}$
$2 n π + \frac{π}{6}$
$2 n π - \frac{7 π}{6}$
$None$ of these

The correct answer is: $2 n pi plus fraction numerator 7 pi over denominator 6 end fraction$

$s i n invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses plus c o s invisible function application open parentheses theta plus fraction numerator 5 pi over denominator 6 end fraction close parentheses equals 2$
$text Since end text sin invisible function application open parentheses 5 A plus fraction numerator pi over denominator 6 end fraction close parentheses less than 1 text and end text rns invisible function application open parentheses A plus fraction numerator 5 pi over denominator 6 end fraction close parentheses less than 1 text , therefore end text$
$text Eq. (1) may hold true if end text s i n invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses text and end text c o s invisible function application open parentheses theta plus fraction numerator 5 pi over denominator 6 end fraction close parentheses text . end text$
$text Both are equal to end text 1 text simultaneously. First, a common value of end text theta text is end text 7 pi divided by 6 text for which end text$
$s i n invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses equals s i n invisible function application fraction numerator 5 pi over denominator 2 end fraction equals s i n invisible function application fraction numerator pi over denominator 2 end fraction equals 1$
and $c o s invisible function application open parentheses theta plus fraction numerator 5 pi over denominator 6 end fraction close parentheses equals c o s invisible function application open parentheses fraction numerator 7 pi over denominator 6 end fraction plus fraction numerator 5 pi over denominator 6 end fraction close parentheses equals c o s invisible function application 2 pi equals 1$
$text Since periodicity of end text s i n invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses text is end text pi text and periodicity of end text$
$c o s invisible function application open parentheses theta plus fraction numerator 5 pi over denominator 6 end fraction close parentheses text is end text 2 pi subscript i end subscript text therefore, the perio end text text dicity end text text of end text s i n invisible function application open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses$
$text in end text open parentheses 2 theta plus fraction numerator pi over denominator 6 end fraction close parentheses text is end text 2 pi$
Therefore, the general solution is
$theta equals 2 n pi plus fraction numerator 7 pi over denominator 6 end fraction$

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