If 0 ≤ x ≤ (π)/(2) , then what is the number of values of x...

The given equation is $\tan x + \sec x = 2\cos x$.

Rewriting the equation in terms of $\sin x and \cos x$:

$\dfrac{\sin x}{\cos x} + \dfrac{1}{\cos x} = 2\cos x$

For the equation to be defined, $\cos x \neq 0$, which means $x \neq \dfrac{\pi}{2}$. Thus, x \in \left[0, \dfrac{\pi}{2}\right).<br><br>Multiplying both sides by\cos x$:<br><br>$\sin x + 1 = 2\cos^2 x$<br><br>Substituting$\cos^2 x = 1 - \sin^2 x$:<br><br>$\sin x + 1 = 2(1 - \sin^2 x)$<br><br>$2\sin^2 x + \sin x - 1 = 0$<br><br>Factoring the quadratic equation:<br><br>$(2\sin x - 1)(\sin x + 1) = 0$<br><br>This gives$\sin x = $\dfrac{1}{2} or \sin x = -1$.

Since $x \in \left[0, \dfrac{\pi}{2}\right)$, $\sin x \geq 0$. Therefore, $\sin x = -1$ is rejected.

For $\sin x = \dfrac{1}{2}$, the only solution in the given interval is $x = \dfrac{\pi}{6}$.

Thus, there is exactly $1$ value of $x$ satisfying the equation.

Answer: $1$