Log-Trigo Equation!!

You have to be careful with this one .....

Note first that the base of a logarithm must be a positive real number other than $1$ . Now $\sqrt{2}\sin(x) \gt 0$ only on the subinterval $(0,\pi)$ , and on this subinterval $\sqrt{2}\sin(x) = 1$ for $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$ . So we are thus restricted to finding solutions $x$ on the subinterval $(0, \pi), x \ne \frac{\pi}{4}, x \ne \frac{3\pi}{4}$ , (A).

Next, the given equation is equivalent to

$(\sqrt{2}\sin(x))^{2} = 1 + \cos(x)$

$\Longrightarrow 2*(1 - \cos^{2}(x)) = 1 + \cos(x)$

$\Longrightarrow 2*(1 - \cos(x))(1 + \cos(x)) = 1 + \cos(x)$ .

Now $1 + \cos(x) = 0$ for $x = \pi$ , which is not a potential solution as it is not on subinterval (A). Thus we can divide through by $(1 + \cos(x))$ to end up with the equation

$2*(1 - \cos(x)) = 1 \Longrightarrow \cos(x) = \frac{1}{2}$ ,

which has only one solution on subinterval (A), namely $x = \frac{\pi}{3}$ .

Thus the correct answer is $\boxed{1}$ .