Trigonometry Problem Analysis

We are given two trigonometric equations involving angles x and y:

• $\sin(x - y) = \frac{\sqrt{3}}{2}$
• $\cos(x + y) = \frac{1}{2}$

We are also given that x and y are positive acute angles ($0^\circ < x < 90^\circ$, $0^\circ < y < 90^\circ$) and $x \ge y$. We need to find the value of x.

Solving Sine Equation for Angle Difference

From the first equation, $\sin(x - y) = \frac{\sqrt{3}}{2}$. Since x and y are positive acute angles and $x \ge y$, the angle difference $(x - y)$ must be in the range $0^\circ \le (x - y) < 90^\circ$. The angle in this range whose sine is $\frac{\sqrt{3}}{2}$ is $60^\circ$.

Therefore, we have:

$x - y = 60^\circ \quad (\text{Equation 1})$

Solving Cosine Equation for Angle Sum

From the second equation, $\cos(x + y) = \frac{1}{2}$. Since x and y are positive acute angles, their sum $(x + y)$ must be in the range $0^\circ < (x + y) < 180^\circ$. The angle in this range whose cosine is $\frac{1}{2}$ is $60^\circ$.

Therefore, we have:

$x + y = 60^\circ \quad (\text{Equation 2})$

Finding Value of x from System of Equations

Now we have a system of two linear equations with two variables x and y:

1. $x - y = 60^\circ$
2. $x + y = 60^\circ$

To find x, we can add Equation 1 and Equation 2:

$ (x - y) + (x + y) = 60^\circ + 60^\circ $

$ 2x = 120^\circ $

Divide by 2:

$ x = \frac{120^\circ}{2} $

$ x = 60^\circ $

The value of x is $60^\circ$.