# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/13) | 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 Property 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 Property 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;  