A curve C has equation y=e^{2x} an x, ext{ where } x eq (2n+1) rac{ ext{π}}{2} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 6

To find the turning points of the curve, we first need to differentiate the given equation. We have:

$y = e^{2x} an x$

Differentiating using the product rule:

$\frac{dy}{dx} = \frac{d}{dx}(e^{2x}) \cdot \tan x + e^{2x} \cdot \frac{d}{dx}( an x)$
$= 2e^{2x} an x + e^{2x} \sec^2 x$

Setting the derivative equal to zero for turning points:

$0 = 2e^{2x} an x + e^{2x} \sec^2 x$

This can be simplified to:

$2 \tan x + \sec^2 x = 0$

From the identity for the secant function:

$\sec^2 x = 1 + \tan^2 x$

Substituting into our equation gives:

$2 \tan x + 1 + \tan^2 x = 0$

This is a quadratic equation in terms of $\tan x$:

$\tan^2 x + 2 \tan x + 1 = 0$

Factoring gives:

$(\tan x + 1)^2 = 0$

From this, we find:

$\tan x = -1$

Thus, the turning points occur where $\tan x = -1$, as required.