1. Let the magnitudes of the two vectors be equal: $|\vec A|=|\vec B|=a$ Let the angle between them be $\theta$.

2. Use the formula for magnitude of sum and difference: $|\vec A+\vec B|^2=|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta$ $|\vec A-\vec B|^2=|\vec A|^2+|\vec B|^2-2|\vec A||\vec B|\cos\theta$

   Since $|\vec A|=|\vec B|=a$, $|\vec A+\vec B|^2=2a^2(1+\cos\theta)$ $|\vec A-\vec B|^2=2a^2(1-\cos\theta)$

3. Given condition: $|\vec A+\vec B|=2|\vec A-\vec B|$

   Squaring both sides, $|\vec A+\vec B|^2=4|\vec A-\vec B|^2$

   Substitute the expressions: $2a^2(1+\cos\theta)=4\cdot 2a^2(1-\cos\theta)$

4. Cancel $2a^2$ from both sides: $1+\cos\theta=4(1-\cos\theta)$

   $1+\cos\theta=4-4\cos\theta$

   $5\cos\theta=3$

   $\cos\theta=\frac{3}{5}$

5. Therefore, $\theta=\cos^{-1}\left(\frac{3}{5}\right)$

6. Checking options:

   • A: $\sin^{-1}(3/5)$ ❌
   • B: $\sin^{-1}(1/3)$ ❌
   • C: $\cos^{-1}(3/5)$ ✅
   • D: $\cos^{-1}(1/3)$ ❌

Hence, the correct answer is Option C.