Exercise 11: Find all real numbers $x$ and $y$ in the interval $[0, 2\pi)$ such that $\begin{cases} \sin x + \sin y = \frac{3}{2} \\ \sin x \sin y = \frac{1}{2} \end{cases}$ Exercise 12: 1. Express $\cos(3x)$ in terms of $\cos x$ and $\sin(3x)$ in terms of $\sin x$.

1. Problem Description

Exercise 11: Find all real numbers $x$ and $y$ in the interval $[0, 2\pi)$ such that

$\begin{cases} \sin x + \sin y = \frac{3}{2} \\ \sin x \sin y = \frac{1}{2} \end{cases}$

Exercise 12:

1. Express $\cos(3x)$ in terms of $\cos x$ and $\sin(3x)$ in terms of $\sin x$.

2. Prove that $\tan(3x) = \tan x \cdot \frac{3 - \tan^2 x}{1 - 3\tan^2 x}$.

Exercise 13:

1. Calculate $(\sqrt{3} + \sqrt{2})^2$.

2. Solve the equation $4x^2 + 2x(\sqrt{3} - \sqrt{2}) - \sqrt{6} = 0$ in $\mathbb{R}$.

3. Deduce the solutions in $[0, 2\pi)$ of the equation $-4\sin^2 y + 2(\sqrt{3} + \sqrt{2}) \cos y - \sqrt{6} + 4 = 0$ and place the images of the solutions of this equation on the trigonometric circle.

2. Solution Steps

Exercise 11:

Let $u = \sin x$ and $v = \sin y$. We have

$\begin{cases} u + v = \frac{3}{2} \\ uv = \frac{1}{2} \end{cases}$

Then $u$ and $v$ are the roots of the quadratic equation

$t^2 - (u+v)t + uv = 0$, which is $t^2 - \frac{3}{2}t + \frac{1}{2} = 0$.

Multiplying by 2, we get $2t^2 - 3t + 1 = 0$.

Then $(2t - 1)(t - 1) = 0$, so $t = 1$ or $t = \frac{1}{2}$.

Thus, $\sin x = 1$ and $\sin y = \frac{1}{2}$, or $\sin x = \frac{1}{2}$ and $\sin y = 1$.

If $\sin x = 1$, then $x = \frac{\pi}{2}$. If $\sin y = \frac{1}{2}$, then $y = \frac{\pi}{6}$ or $y = \frac{5\pi}{6}$.

If $\sin x = \frac{1}{2}$, then $x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$. If $\sin y = 1$, then $y = \frac{\pi}{2}$.

Therefore, the solutions are $(x, y) = (\frac{\pi}{2}, \frac{\pi}{6})$, $(x, y) = (\frac{\pi}{2}, \frac{5\pi}{6})$, $(x, y) = (\frac{\pi}{6}, \frac{\pi}{2})$, $(x, y) = (\frac{5\pi}{6}, \frac{\pi}{2})$.

Exercise 12:

1. $\cos(3x) = \cos(2x + x) = \cos(2x) \cos x - \sin(2x) \sin x = (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x = 2\cos^3 x - \cos x - 2\sin^2 x \cos x = 2\cos^3 x - \cos x - 2(1 - \cos^2 x) \cos x = 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x$.

$\sin(3x) = \sin(2x + x) = \sin(2x)\cos x + \cos(2x)\sin x = (2\sin x \cos x)\cos x + (1 - 2\sin^2 x)\sin x = 2\sin x \cos^2 x + \sin x - 2\sin^3 x = 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x$.

Thus $\cos(3x) = 4\cos^3 x - 3\cos x$ and $\sin(3x) = 3\sin x - 4\sin^3 x$.

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

However, the problem statement suggests a different approach.

$\tan(3x) = \frac{\sin(3x)}{\cos(3x)} = \frac{3\sin x - 4\sin^3 x}{4\cos^3 x - 3\cos x} = \frac{\frac{3\sin x}{\cos x} - 4\frac{\sin^3 x}{\cos^3 x}}{4\frac{\cos^3 x}{\cos x} - \frac{3\cos x}{\cos x}} = \frac{3\tan x - 4\tan^3 x/\cos^2 x}{4\cos^2x - 3}$. This doesn't seem to go anywhere.

$\tan(3x) = \tan(2x + x) = \frac{\tan(2x) + \tan x}{1 - \tan(2x)\tan x}$.

$\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$.

$\tan(3x) = \frac{\frac{2\tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2\tan x}{1 - \tan^2 x}\tan x} = \frac{2\tan x + \tan x(1 - \tan^2 x)}{1 - \tan^2 x} \times \frac{1 - \tan^2 x}{1 - \tan^2 x - 2\tan^2 x} = \frac{2\tan x + \tan x - \tan^3 x}{1 - 3\tan^2 x} = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = \tan x \frac{3 - \tan^2 x}{1 - 3\tan^2 x}$.

Exercise 13:

1. $(\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$.

2. $4x^2 + 2x(\sqrt{3} - \sqrt{2}) - \sqrt{6} = 0$.

$x = \frac{-2(\sqrt{3} - \sqrt{2}) \pm \sqrt{4(\sqrt{3} - \sqrt{2})^2 - 4(4)(-\sqrt{6})}}{8} = \frac{-(\sqrt{3} - \sqrt{2}) \pm \sqrt{(\sqrt{3} - \sqrt{2})^2 + 4\sqrt{6}}}{4}$.

$(\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}$.

$x = \frac{-(\sqrt{3} - \sqrt{2}) \pm \sqrt{5 - 2\sqrt{6} + 4\sqrt{6}}}{4} = \frac{-(\sqrt{3} - \sqrt{2}) \pm \sqrt{5 + 2\sqrt{6}}}{4} = \frac{-(\sqrt{3} - \sqrt{2}) \pm (\sqrt{3} + \sqrt{2})}{4}$.

$x_1 = \frac{-\sqrt{3} + \sqrt{2} + \sqrt{3} + \sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$.

$x_2 = \frac{-\sqrt{3} + \sqrt{2} - \sqrt{3} - \sqrt{2}}{4} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$.

3. $-4\sin^2 y + 2(\sqrt{3} + \sqrt{2}) \cos y - \sqrt{6} + 4 = 0$.

$-4(1 - \cos^2 y) + 2(\sqrt{3} + \sqrt{2}) \cos y - \sqrt{6} + 4 = 0$.

$4\cos^2 y + 2(\sqrt{3} + \sqrt{2}) \cos y - \sqrt{6} = 0$.

Let $x = \cos y$.

$4x^2 + 2(\sqrt{3} + \sqrt{2})x - \sqrt{6} = 0$. This looks similar to the previous equation, but the middle term has the wrong sign.

However $4x^2 + 2(\sqrt{3}+\sqrt{2})x - \sqrt{6}=0$ implies $x=\frac{-\sqrt{3}-\sqrt{2} \pm (\sqrt{3}-\sqrt{2})}{4}$.

$x_1 = \frac{-\sqrt{3}-\sqrt{2}+\sqrt{3}-\sqrt{2}}{4} = -\frac{2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$.

$x_2 = \frac{-\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}}{4} = -\frac{2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$.

$\cos y = -\frac{\sqrt{2}}{2}$, so $y = \frac{3\pi}{4}$ or $y = \frac{5\pi}{4}$.

$\cos y = -\frac{\sqrt{3}}{2}$, so $y = \frac{5\pi}{6}$ or $y = \frac{7\pi}{6}$.

3. Final Answer

Exercise 11: $(x, y) = (\frac{\pi}{2}, \frac{\pi}{6})$, $(x, y) = (\frac{\pi}{2}, \frac{5\pi}{6})$, $(x, y) = (\frac{\pi}{6}, \frac{\pi}{2})$, $(x, y) = (\frac{5\pi}{6}, \frac{\pi}{2})$.

Exercise 12:

1. $\cos(3x) = 4\cos^3 x - 3\cos x$ and $\sin(3x) = 3\sin x - 4\sin^3 x$.

2. $\tan(3x) = \tan x \frac{3 - \tan^2 x}{1 - 3\tan^2 x}$.

Exercise 13:

1. $(\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6}$.

2. $x = \frac{\sqrt{2}}{2}$ and $x = -\frac{\sqrt{3}}{2}$.

3. $y = \frac{3\pi}{4}$, $y = \frac{5\pi}{4}$, $y = \frac{5\pi}{6}$, $y = \frac{7\pi}{6}$.