f(x)= 2secx + tanx for all x in the interval 0 and 2pi
Am i supposed to take the derivative of f(x), set it equal to 0, and find all points where x satisfies the equation?
Factor out a $\displaystyle \sec{x}$
$\displaystyle \sec{x}(2\tan{x} + \sec{x})= 0$
so now you have
$\displaystyle \sec{x} = 0$ and $\displaystyle 2\tan{x} + \sec{x} =0$ and you don't have to worry about sec(x) = 0 since that is not possible
$\displaystyle \frac{2\sin{x}}{\cos{x}}+ \frac{1}{\cos{x}} = 0$
$\displaystyle \frac{2\sin{x} + 1}{\cos{x}}$
$\displaystyle 2\sin{x} + 1 = 0$
$\displaystyle 2\sin{x} = -1$
$\displaystyle \sin{x} = -\frac{1}{2}$
$\displaystyle x= \frac{7\pi}{6}$ and $\displaystyle x= \frac{11\pi}{6}$
Unit Circle
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