To find where \( f(x) = \sin 3x - \cos 3x \) is strictly increasing, we examine where its derivative is positive.
Step 1: Compute the derivative of \( f(x) \). Differentiating \( f(x) \) with respect to \( x \) yields: \[ f'(x) = \frac{d}{dx} \left( \sin 3x - \cos 3x \right) \] Applying the chain rule: \[ f'(x) = 3\cos 3x + 3\sin 3x \] Factoring out 3: \[ f'(x) = 3\left( \cos 3x + \sin 3x \right) \] Step 2: Establish the inequality \( f'(x)>0 \) to identify intervals of increase. We require \( f'(x)>0 \): \[ 3\left( \cos 3x + \sin 3x \right)>0 \quad \Rightarrow \quad \cos 3x + \sin 3x>0 \] Rewrite \( \cos 3x + \sin 3x \) using the identity \( \cos \theta + \sin \theta = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \): \[ \cos 3x + \sin 3x = \sqrt{2} \sin \left( 3x + \frac{\pi}{4} \right) \] The inequality transforms to: \[ \sqrt{2} \sin \left( 3x + \frac{\pi}{4} \right)>0 \] Dividing by \( \sqrt{2} \): \[ \sin \left( 3x + \frac{\pi}{4} \right)>0 \] Step 3: Solve the trigonometric inequality. Since \( \sin \theta>0 \) for \( \theta \in \left( 0, \pi \right) \), we have: \[ 0<3x + \frac{\pi}{4}<\pi \] Isolate \( 3x \) by subtracting \( \frac{\pi}{4} \): \[ -\frac{\pi}{4}<3x<\frac{3\pi}{4} \] Divide by 3: \[ -\frac{\pi}{12}<x<\frac{\pi}{4} \] The function is strictly increasing in the interval \( \left( -\frac{\pi}{12}, \frac{\pi}{4} \right) \). Step 4: Determine the final intervals of increase within the given domain \( 0<x<\frac{\pi}{2} \). The function \( f(x) \) is strictly increasing in the union of intervals: \[ x \in \left( 0, \frac{\pi}{4} \right) \]