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;

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