Q. Solve \( \arcsin\left(\frac{x}{3} + 5\right) = \arcsin\left(2x + 30\right) \).

Solution

1. Write the equation:
   \( \arcsin\!\left(\dfrac{x}{3}+5\right)=\arcsin(2x+30) \)
2. Impose domain restrictions for both arcsine arguments (they must lie in \([-1,1]\)):

   For the left argument:
   \( -1 \le \dfrac{x}{3}+5 \le 1 \)
   Subtract 5 from all parts:
   \( -6 \le \dfrac{x}{3} \le -4 \)
   Multiply by 3:
   \( -18 \le x \le -12 \)

   For the right argument:
   \( -1 \le 2x+30 \le 1 \)
   Subtract 30:
   \( -31 \le 2x \le -29 \)
   Divide by 2:
   \( -\dfrac{31}{2} \le x \le -\dfrac{29}{2} \)

   The solution must satisfy both ranges, so
   \( x \in \left[-\dfrac{31}{2},-\dfrac{29}{2}\right] \) (i.e. \([-15.5,-14.5]\)).

3. Use monotonicity of arcsine: on \([-1,1]\) the function \(\arcsin\) is strictly increasing (its derivative \(1/\sqrt{1-u^{2}}\) is positive for \(|u|<1\)). Therefore if \(\arcsin(A)=\arcsin(B)\) with \(A,B\in[-1,1]\) then \(A=B\). Set the arguments equal: \( \dfrac{x}{3}+5 = 2x+30 \)
4. Solve the linear equation:
   Multiply both sides by 3:
   \( x+15 = 6x+90 \)
   Bring terms together:
   \( 15-90 = 6x-x \)
   \( -75 = 5x \)
   Divide by 5:
   \( x = -15 \)
5. Verify the solution lies in the domain and satisfies the original equation:
   \( \dfrac{x}{3}+5 = \dfrac{-15}{3}+5 = -5+5 = 0 \)
   \( 2x+30 = 2(-15)+30 = -30+30 = 0 \)
   Both arguments equal 0 which is in \([-1,1]\), and \(\arcsin(0)=0\), so the equation is satisfied.
6. Conclusion:
   The unique solution is \( \boxed{x=-15} \).