In the diagram, the graphs of $f(x)= ext{sin}x-1$ and $g(x)= ext{cos}2x$ are drawn for the interval $x \in [-90^{\circ};360^{\circ}]$ - NSC Mathematics - Question 6 - 2019 - Paper 2

To find where the graph $f$ is decreasing, we need to find the intervals where its derivative is negative. The derivative of $f(x)= ext{sin}x-1$ is $f'(x)= ext{cos}x$.

Thus, $f$ is decreasing when:

$\text{cos}x < 0$

This occurs in the intervals $x \in (90^{\circ}; 270^{\circ})$ on the interval $x \in [-90^{\circ}; 360^{\circ}]$, which leads to the values:

$x \in [90^{\circ}; 270^{\circ}]$

Since $PQ$ is parallel to the $y$-axis, the $x$-coordinates of points $P$ and $Q$ will be the same. Therefore, we set the equations equal:

$PQ = ext{cos}2x - (\text{sin}x - 1)$

This simplifies to:

$PQ = \text{cos}2x - \text{sin}x + 1$

To find the maximum of this, we can take its derivative and set it to zero:

First, we calculate:

$\frac{d}{dx}(\text{cos}2x - \text{sin}x + 1) = -2\text{sin}2x - \text{cos}x$

Setting this equal to zero yields a critical evaluation leading us to:

$\text{sin}x = \frac{1}{4}$

The solutions are: $x = 194.48^{\circ} \text{ or } 345.52^{\circ}$