You're on a roll — Step 1: Expand the right side of the equation using the compound angle formula for cosine. The given equation is $7\cos 2x - 24\sin 2x = R\cos(2x+\alpha)$. Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we expand $R\cos(2x+\alpha)$: $$ R\cos(2x+\alpha) = R(\cos 2x \cos \alpha - \sin 2x \sin \alpha) $$ $$ R\cos(2x+\alpha) = (R\cos\alpha)\cos 2x - (R\sin\alpha)\sin 2x $$ Step 2: Compare the coefficients of $\cos 2x$ and $\sin 2x$. By comparing the expanded form with the left side of the given equation, $7\cos 2x - 24\sin 2x$: 1) $R\cos\alpha = 7$ 2) $R\sin\alpha = 24$ (Note: The negative sign in front of $24\sin 2x$ matches the negative sign in the expanded formula, so $R\sin\alpha$ is positive.) Step 3: Determine the value of $R$. Square both equations from Step 2 and add them: $$ (R\cos\alpha)^2 + (R\sin\alpha)^2 = 7^2 + 24^2 $$ $$ R^2\cos^2\alpha + R^2\sin^2\alpha = 49 + 576 $$ $$ R^2(\cos^2\alpha + \sin^2\alpha) = 625 $$ Since $\cos^2\alpha + \sin^2\alpha = 1$: $$ R^2 = 625 $$ Given that $R > 0$: $$ R = \sqrt{625} $$ $$ R = 25 $$ Step 4: Determine the value of $\alpha$. Divide equation (2) by equation (1): $$ \frac{R\sin\alpha}{R\cos\alpha} = \frac{24}{7} $$ $$ \tan\alpha = \frac{24}{7} $$ Given that $0 < \alpha < 90^\circ$, $\alpha$ is in the first quadrant. $$ \alpha = \arctan\left(\frac{24}{7}\right) $$ Using a calculator: $$ \alpha \approx 73.73979^\circ $$ Rounding to two decimal places: $$ \alpha \approx 73.74^\circ $$ The values are $\boxed{R=25, \alpha \approx 73.74^\circ}$. Send me the next one 📸