Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2011  OctNov  (P19709/13)  Q#5
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Question
i. Given that
Show that, for real values of
ii. Hence solve the equation
for .
Solution
i.
We have the equation;
We have the trigonometric identity;
We can rewrite it as;
Thus
Now we have two options;









NOT POSSIBLE 
So we are left with ONLY option;
ii.
To solve the equation , for , as demonstrated in (i), we can write the given equation as;
To solve this equation, we can substitute . Hence,
Since given interval is , for interval can be found as follows;
Adding on both sides of the inequality;
Since ;
Hence the given interval for is .
To solve equation for interval ,
Using calculator we can find the values of .
We utilize the symmetry property of to find other solutions (roots) of :


Symmetry 

Hence;
For
Therefore, we have two solutions (roots) of the equation;


To find all the solutions (roots) of we utilize the periodic property of .


Periodic 
or

Therefore;




For 
For 




For















Hence only following solutions (roots) are within the given interval ;


Since ;






Hence, all the solutions of the equation , within the interval
for are;


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