\( \sqrt { 2 } \sin ( x + a 5 ^ { \circ } ) = \sin x + \cos x \)

To solve the equation: \[ \sqrt{2} \sin(x + a \cdot 5^\circ) = \sin x + \cos x \] we can proceed as follows: ### Step 1: Simplify the Right Side First, recognize that the right side of the equation can be expressed using a single sine function with a phase shift. Recall that: \[ \sin x + \cos x = \sqrt{2} \sin\left(x + 45^\circ\right) \] This is derived from the identity: \[ \sin x + \cos x = \sqrt{2} \sin\left(x + 45^\circ\right) \] ### Step 2: Equate Both Sides Now, substitute the simplified right side back into the original equation: \[ \sqrt{2} \sin(x + a \cdot 5^\circ) = \sqrt{2} \sin\left(x + 45^\circ\right) \] Since the \(\sqrt{2}\) terms cancel out, we have: \[ \sin(x + a \cdot 5^\circ) = \sin\left(x + 45^\circ\right) \] ### Step 3: Solve for the Phase Shift The equation \(\sin(A) = \sin(B)\) implies that: \[ A = B + 360^\circ \cdot k \quad \text{or} \quad A = 180^\circ - B + 360^\circ \cdot k \quad \text{for any integer } k \] Applying this to our equation: 1. **First Case:** \[ x + a \cdot 5^\circ = x + 45^\circ + 360^\circ \cdot k \] Simplifying: \[ a \cdot 5^\circ = 45^\circ + 360^\circ \cdot k \] \[ a = \frac{45^\circ + 360^\circ \cdot k}{5^\circ} = 9 + 72 \cdot k \] 2. **Second Case:** \[ x + a \cdot 5^\circ = 180^\circ - (x + 45^\circ) + 360^\circ \cdot k \] \[ x + a \cdot 5^\circ = 135^\circ - x + 360^\circ \cdot k \] \[ 2x + a \cdot 5^\circ = 135^\circ + 360^\circ \cdot k \] \[ x = \frac{135^\circ - a \cdot 5^\circ + 360^\circ \cdot k}{2} \] However, this case imposes a condition on \(x\), making it less general unless \(a\) is chosen to satisfy the equation for all \(x\). ### Conclusion The most straightforward solution comes from the first case, leading to: \[ a = 9 + 72 \cdot k \] where \(k\) is any integer. Typically, the smallest positive value is taken, so: \[ a = 9^\circ \] **Answer:** The constant a must equal 9°. Thus, the solution is a = 9°.