Let's solve each of the problems one by one:

Problem 14: $2 \sin^2 x - \sin x - 1 = 0$

This equation is a trigonometric quadratic equation in terms of $\sin x$. Let $u = \sin x$, so the equation becomes:

$2u^2 - u - 1 = 0$

Step 1: Solving the Quadratic Equation

This can be factored or solved using the quadratic formula. Using the quadratic formula:

$u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$

This gives us two solutions for $u$:

1. $u = \frac{1 + 3}{4} = 1$
2. $u = \frac{1 - 3}{4} = -\frac{1}{2}$

Since $u = \sin x$, we have:

1. $\sin x = 1$
2. $\sin x = -\frac{1}{2}$

Step 2: Finding All Solutions

1. For $\sin x = 1$:

   • $x = \frac{\pi}{2} + 2\pi k$ for any integer $k$.

2. For $\sin x = -\frac{1}{2}$:

   • The solutions for $x$ are $x = \frac{7\pi}{6} + 2\pi k$ and $x = \frac{11\pi}{6} + 2\pi k$, where $k$ is any integer.

So, the general solutions for all values of $x$ are:

$x = \frac{\pi}{2} + 2\pi k, \quad x = \frac{7\pi}{6} + 2\pi k, \quad x = \frac{11\pi}{6} + 2\pi k$

Step 3: Finding Solutions in the Interval $0 \leq x < 2\pi$

For $0 \leq x < 2\pi$, we only consider:

$x = \frac{\pi}{2}, \quad x = \frac{7\pi}{6}, \quad x = \frac{11\pi}{6}$

Problem 15: $\sqrt{3} \tan \theta - 2 \sin \theta \cos \theta = 0$

Rewrite the second term using the double-angle identity: $\sin 2\theta = 2 \sin \theta \cos \theta$. The equation becomes:

$\sqrt{3} \tan \theta - \sin 2\theta = 0$

Dividing both sides by $\cos \theta$, we get:

$\sqrt{3} - \sin 2\theta = 0$

Let’s proceed with