3 Calculate the value of sin 2A if cos A = 24/25 and A is in the first quadrant. 3.4 Solve for x if: 3 cos x tan x = tan x Where, 0° ≤ x ≤ 360° 3.5 Simplify: cos(315) sin(210) tan(225) / sin(-45) cos(240)

Since $A$ is in the first quadrant, $\sin A > 0$. We use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$.

The value of $\sin 2A$ is $\boxed{\frac{336}{625}}$.

Case 1: $\tan x = 0$
This occurs when $\sin x = 0$.

(Note: $\tan x$ is undefined at $90^\circ$ and $270^\circ$, but these are not solutions for $\tan x = 0$.)

Case 2: $3 \cos x - 1 = 0$

The reference angle is $\alpha = \arccos\left(\frac{1}{3}\right) \approx 70.53^\circ$.
Since $\cos x$ is positive, $x$ is in Quadrant I or Quadrant IV.
In Quadrant I: $x_1 \approx 70.53^\circ$
In Quadrant IV: $x_2 = 360^\circ - \alpha \approx 360^\circ - 70.53^\circ \approx 289.47^\circ$

The solutions for $x$ are approximately $\boxed{0^\circ, 70.53^\circ, 180^\circ, 289.47^\circ, 360^\circ}$.

• $\cos(315^\circ) = \cos(360^\circ - 45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$
• $\sin(210^\circ) = \sin(180^\circ + 30^\circ) = -\sin(30^\circ) = -\frac{1}{2}$
• $\tan(225^\circ) = \tan(180^\circ + 45^\circ) = \tan(45^\circ) = 1$
• $\sin(-45^\circ) = -\sin(45^\circ) = -\frac{\sqrt{2}}{2}$
• $\cos(240^\circ) = \cos(180^\circ + 60^\circ) = -\cos(60^\circ) = -\frac{1}{2}$

The simplified value is $\boxed{-1}$.