In the diagram, the graph of $f(x)= ext{cos}2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}},90^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2017 - Paper 2

To find the intersection point $A$:

1. Set $f(x)$ equal to $g(x)$:

   $ext{cos}2x = 2 ext{sin}x - 1.$

2. Rearranging: We can convert $ext{cos}2x$ using the double angle identity:

   $ext{cos}2x = 1 - 2 ext{sin}^2x$

   Thus,

   $1 - 2 ext{sin}^2x = 2 ext{sin}x - 1.$

3. Reorganizing into a standard quadratic form:

   $2 ext{sin}^2x + 2 ext{sin}x - 2 = 0.$

4. Simplifying further gives:

   $ext{sin}^2x + ext{sin}x - 1 = 0.$

5. Apply the quadratic formula:

   $b = 1, a = 1, c = -1$

   ext{sin}x = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a} = rac{-1 ext{±} ext{√}(1^2 - 4(1)(-1))}{2(1)}

   This simplifies to:

   ext{sin}x = rac{-1 ext{±} ext{√}5}{2}.

6. Select valid solutions: Since $ext{sin}x$ can only be between -1 and 1, the valid solution is:

   ext{sin}x = rac{-1+ ext{√}5}{2}

Now we will find the coordinates of the intersection points:

1. Identify the valid solutions for $x$: From the previous part, we found:

   ext{sin}x = rac{-1+ ext{√}5}{2} ext{ which approximates to } 0.618.

2. Calculate $x$ values: Using reference angles:

   $x ext{ can be } 38.17^{ ext{o}} ext{ (1st quadrant)} ext{ or } x = 141.83^{ ext{o}} ext{ (2nd quadrant)}$

3. Find corresponding $y$ values: Substituting back into either $f(x)$ or $g(x)$ will yield:

   • At $x = 38.17^{ ext{o}}$: $y = f(38.17^{ ext{o}}) ext{ gives } y ext{ approximately } 0.24$
   • At $x = 141.83^{ ext{o}}$: $y = f(141.83^{ ext{o}}) ext{ gives the same } y ext{ value } 0.24$

4. Final coordinates: Thus, the coordinates of intersection points are:

   • $(38.17^{ ext{o}}, 0.24)$
   • $(141.83^{ ext{o}}, 0.24)$