We are given two exercises. Exercise 16: We are given the equation (E): $8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3} = 0$. 1. We need to verify that $\frac{\sqrt{3}}{2}$ is a solution of the equation (E).

1. Problem Description

We are given two exercises.

Exercise 16:

We are given the equation (E): $8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3} = 0$.

1. We need to verify that $\frac{\sqrt{3}}{2}$ is a solution of the equation (E).

2. We need to find all the solutions of (E).

3. We need to solve in $\mathbb{R}$ the equation: $8\sin^3 x - 4\sqrt{3}\sin^2 x - 2\sin x + \sqrt{3} = 0$.

Exercise 17:

We are given that $x$ and $y$ are elements of $[0; \frac{\pi}{2}]$ such that $\sin x = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\cos y = \frac{\sqrt{3}}{2}$.

1. We need to verify that $(\frac{\sqrt{6}+\sqrt{2}}{4})^2 = \frac{2-\sqrt{3}}{4}$.

2. We need to calculate $\cos x$ and $\sin y$. What is the value of $y$?

2. Solution Steps

Exercise 16:

1. To verify that $\frac{\sqrt{3}}{2}$ is a solution, we substitute $x = \frac{\sqrt{3}}{2}$ into the equation (E):

$8(\frac{\sqrt{3}}{2})^3 - 4\sqrt{3}(\frac{\sqrt{3}}{2})^2 - 2(\frac{\sqrt{3}}{2}) + \sqrt{3} = 8(\frac{3\sqrt{3}}{8}) - 4\sqrt{3}(\frac{3}{4}) - \sqrt{3} + \sqrt{3} = 3\sqrt{3} - 3\sqrt{3} - \sqrt{3} + \sqrt{3} = 0$.

Therefore, $x = \frac{\sqrt{3}}{2}$ is a solution.

2. Since $x = \frac{\sqrt{3}}{2}$ is a solution, we can factor $(x - \frac{\sqrt{3}}{2})$ from the polynomial $8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3}$.

Let's divide the polynomial by $(x - \frac{\sqrt{3}}{2})$ or equivalently by $(2x - \sqrt{3})$.

$8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3} = (2x - \sqrt{3})(4x^2 - 1)$.

Therefore, $8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3} = 0$ if and only if $(2x - \sqrt{3})(4x^2 - 1) = 0$.

So, $2x - \sqrt{3} = 0$ or $4x^2 - 1 = 0$.

$2x = \sqrt{3}$, so $x = \frac{\sqrt{3}}{2}$.

$4x^2 = 1$, so $x^2 = \frac{1}{4}$, which means $x = \pm \frac{1}{2}$.

Thus, the solutions are $x = \frac{\sqrt{3}}{2}$, $x = \frac{1}{2}$, and $x = -\frac{1}{2}$.

3. The equation $8\sin^3 x - 4\sqrt{3}\sin^2 x - 2\sin x + \sqrt{3} = 0$ is the same as $8x^3 - 4\sqrt{3}x^2 - 2x + \sqrt{3} = 0$ with $x = \sin x$.

The solutions are $\sin x = \frac{\sqrt{3}}{2}$, $\sin x = \frac{1}{2}$, and $\sin x = -\frac{1}{2}$.

If $\sin x = \frac{\sqrt{3}}{2}$, then $x = \frac{\pi}{3} + 2k\pi$ or $x = \frac{2\pi}{3} + 2k\pi$, $k \in \mathbb{Z}$.

If $\sin x = \frac{1}{2}$, then $x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{5\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.

If $\sin x = -\frac{1}{2}$, then $x = -\frac{\pi}{6} + 2k\pi$ or $x = \frac{7\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.

Exercise 17:

1. $(\frac{\sqrt{6}+\sqrt{2}}{4})^2 = \frac{(\sqrt{6})^2 + 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2}{16} = \frac{6 + 2\sqrt{12} + 2}{16} = \frac{8 + 2(2\sqrt{3})}{16} = \frac{8 + 4\sqrt{3}}{16} = \frac{2 + \sqrt{3}}{4}$.

This contradicts the equation $(\frac{\sqrt{6}+\sqrt{2}}{4})^2 = \frac{2-\sqrt{3}}{4}$. There appears to be a sign error.

Let's verify $(\frac{\sqrt{6}-\sqrt{2}}{4})^2 = \frac{6 - 2\sqrt{12} + 2}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$.

Thus, it should be $(\frac{\sqrt{6}-\sqrt{2}}{4})^2 = \frac{2-\sqrt{3}}{4}$. However, the question states $(\frac{\sqrt{6}+\sqrt{2}}{4})^2 = \frac{2-\sqrt{3}}{4}$.

2. Since $\sin x = \frac{\sqrt{6}+\sqrt{2}}{4}$, we use the identity $\sin^2 x + \cos^2 x = 1$ to find $\cos x$.

$\cos^2 x = 1 - \sin^2 x = 1 - (\frac{\sqrt{6}+\sqrt{2}}{4})^2 = 1 - \frac{2+\sqrt{3}}{4} = \frac{4 - 2 - \sqrt{3}}{4} = \frac{2 - \sqrt{3}}{4}$.

Therefore, $\cos x = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2-\sqrt{3}}}{2}$.

Since $\sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2}$, $\cos x = \frac{\sqrt{6}-\sqrt{2}}{4}$.

Since $\cos y = \frac{\sqrt{3}}{2}$, and $y \in [0; \frac{\pi}{2}]$, then $y = \arccos(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$.

Since $\sin y = \sin(\frac{\pi}{6}) = \frac{1}{2}$, we have $\sin y = \frac{1}{2}$.

3. Final Answer

Exercise 16:

1. Verified that $x = \frac{\sqrt{3}}{2}$ is a solution.

2. The solutions are $x = \frac{\sqrt{3}}{2}$, $x = \frac{1}{2}$, and $x = -\frac{1}{2}$.

3. The solutions are $x = \frac{\pi}{3} + 2k\pi$, $x = \frac{2\pi}{3} + 2k\pi$, $x = \frac{\pi}{6} + 2k\pi$, $x = \frac{5\pi}{6} + 2k\pi$, $x = -\frac{\pi}{6} + 2k\pi$, and $x = \frac{7\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$.

Exercise 17:

1. $(\frac{\sqrt{6}+\sqrt{2}}{4})^2 = \frac{2 + \sqrt{3}}{4}$, which is NOT equal to $\frac{2 - \sqrt{3}}{4}$.

2. $\cos x = \frac{\sqrt{6}-\sqrt{2}}{4}$, $\sin y = \frac{1}{2}$, and $y = \frac{\pi}{6}$.