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If $A equals open curly brackets theta blank colon cos invisible function application theta greater than negative fraction numerator 1 over denominator 2 end fraction comma 0 less or equal than theta less or equal than pi close curly brackets$ and
$B equals open curly brackets theta blank colon sin invisible function application theta greater than fraction numerator 1 over denominator 2 end fraction comma fraction numerator pi over denominator 3 end fraction less or equal than theta less or equal than pi close curly brackets comma$ then

(a) $A \cap B = {θ : π / 3 \leq θ \leq 2 π / 3}$
$A \cap B = {θ : - π / 3 \leq θ \leq 2 π / 3}$
$A \cup B = {θ : - 5 π / 6 \leq θ \leq 5 π / 6}$
$A \cup B = {θ : 0 \leq θ \leq π / 6}$

The correct answer is: (a) $A intersection B equals left curly bracket theta blank colon pi divided by 3 less or equal than theta less or equal than 2 pi divided by 3 right curly bracket$

We have,
$cos invisible function application theta greater than negative fraction numerator 1 over denominator 2 end fraction blank a n d blank 0 less or equal than theta less or equal than pi$
$rightwards double arrow 0 less or equal than theta less or equal than 2 pi divided by 3$ and $0 less or equal than theta less or equal than pi$
$rightwards double arrow 0 less or equal than theta less or equal than fraction numerator 2 pi over denominator 3 end fraction blank rightwards double arrow A equals left curly bracket theta colon 0 less or equal than theta less or equal than 2 pi divided by 3 right curly bracket$
Also,
$sin invisible function application theta greater than fraction numerator 1 over denominator 2 end fraction blank a n d blank pi divided by 3 less or equal than theta less or equal than pi$
$rightwards double arrow fraction numerator pi over denominator 3 end fraction less or equal than theta less or equal than fraction numerator 5 blank pi over denominator 6 end fraction rightwards double arrow B equals open curly brackets theta colon fraction numerator pi over denominator 3 end fraction less or equal than theta less or equal than fraction numerator 5 blank pi over denominator 6 end fraction close curly brackets$
$therefore A intersection B equals open curly brackets theta colon fraction numerator pi over denominator 3 end fraction less or equal than theta less or equal than fraction numerator 2 blank pi over denominator 3 end fraction close curly brackets blank a n d blank A union B equals open curly brackets theta colon 0 less or equal than theta less or equal than fraction numerator 5 pi over denominator 6 end fraction close curly brackets$

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