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Question

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

$θ = 2 n π \pm \frac{π}{4} + \frac{π}{12}$
$θ = n π + {(- 1)}^{n} \frac{π}{4} + \frac{π}{12}$
$θ = 2 n π \pm \frac{π}{4}$
$θ = n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{12}$

The correct answer is: $theta equals 2 n pi plus-or-minus fraction numerator pi over denominator 4 end fraction plus fraction numerator pi over denominator 12 end fraction$

$open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2$
$rightwards double arrow fraction numerator open parentheses square root of 3 minus 1 close parentheses over denominator 2 square root of 2 end fraction sin invisible function application theta plus open parentheses fraction numerator square root of 3 plus 1 over denominator 2 square root of 2 end fraction close parentheses cos invisible function application theta equals fraction numerator 1 over denominator square root of 2 end fraction$
$rightwards double arrow sin invisible function application fraction numerator pi over denominator 12 end fraction sin invisible function application theta plus cos invisible function application fraction numerator pi over denominator 12 end fraction cos invisible function application theta equals cos invisible function application fraction numerator pi over denominator 4 end fraction$
$rightwards double arrow cos invisible function application open parentheses theta minus fraction numerator pi over denominator 12 end fraction close parentheses equals cos invisible function application fraction numerator pi over denominator 4 end fraction$
$rightwards double arrow theta minus fraction numerator pi over denominator 12 end fraction equals 2 n pi plus-or-minus fraction numerator pi over denominator 4 end fraction comma blank n element of Z$
$equals 2 n pi plus-or-minus fraction numerator pi over denominator 4 end fraction plus fraction numerator pi over denominator 12 end fraction$

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