a) Express $f(x)$ in the form $R \cos(x - \lambda)$. Step 1: Expand $R \cos(x - \lambda)$ using the compound angle formula. $$ R \cos(x - \lambda) = R(\cos x \cos \lambda + \sin x \sin \lambda) $$ $$ R \cos(x - \lambda) = (R \cos \lambda) \cos x + (R \sin \lambda) \sin x $$ Step 2: Compare coefficients with $f(x) = \cos x + \sqrt{3} \sin x$. We have: $R \cos \lambda = 1 \quad (1)$ $R \sin \lambda = \sqrt{3} \quad (2)$ Step 3: Find $R$. Square and add equations (1) and (2): $$ (R \cos \lambda)^2 + (R \sin \lambda)^2 = 1^2 + (\sqrt{3})^2 $$ $$ R^2 \cos^2 \lambda + R^2 \sin^2 \lambda = 1 + 3 $$ $$ R^2 (\cos^2 \lambda + \sin^2 \lambda) = 4 $$ Using the identity $\cos^2 \lambda + \sin^2 \lambda = 1$: $$ R^2 (1) = 4 $$ $$ R^2 = 4 $$ Since $R > 0$, $R = \sqrt{4} = 2$. Step 4: Find $\lambda$. Divide equation (2) by equation (1): $$ \frac{R \sin \lambda}{R \cos \lambda} = \frac{\sqrt{3}}{1} $$ $$ \tan \lambda = \sqrt{3} $$ Given $0 < \lambda < \frac{\pi}{2}$, $\lambda$ is in the first quadrant. $$ \lambda = \arctan(\sqrt{3}) = \frac{\pi}{3} $$ Step 5: Write $f(x)$ in the required form. $$ f(x) = 2 \cos \left(x - \frac{\pi}{3}\right) $$ The expression for $f(x)$ is $\boxed{2 \cos \left(x - \frac{\pi}{3}\right)}$. b) Find the minimum value of $\frac{1}{1+|f(x)|}$. Step 1: Determine the range of $f(x)$. From part (a), $f(x) = 2 \cos \left(x - \frac{\pi}{3}\right)$. The range of $\cos \theta$ is $[-1, 1]$. Therefore, the range of $f(x)$ is $2 \times [-1, 1] = [-2, 2]$. Step 2: Determine the maximum value of $|f(x)|$. Since $f(x)$ ranges from $-2$ to $2$, the maximum value of $|f(x)|$ is $2$. This occurs when $f(x) = 2$ or $f(x) = -2$. Step 3: Find the minimum value of the expression. To minimize the fraction $\frac{1}{1+|f(x)|}$, its denominator $1+|f(x)|$ must be maximized. The maximum value of $|f(x)|$ is $2$. So, the maximum value of $1+|f(x)|$ is $1+2 = 3$. The minimum value of $\frac{1}{1+|f(x)|}$ is $\frac{1}{3}$. The minimum value is $\boxed{\frac{1}{3}}$. c) Find the general solution of $f(x) = \sqrt{3}$. Step 1: Set $f(x)$ equal to $\sqrt{3}$. Using the expression for $f(x)$ from part (a): $$ 2 \cos \left(x - \frac{\pi}{3}\right) = \sqrt{3} $$ $$ \cos \left(x - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $$ Step 2: Find the principal value. Let $Y = x - \frac{\pi}{3}$. We need to solve $\cos Y = \frac{\sqrt{3}}{2}$. The principal value for $Y$ is $\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$. Step 3: Write the general solution for $Y$. The general solution for $\cos Y = \cos \alpha$ is $Y = 2n\pi \pm \alpha$, where $n \in \mathbb{Z}$. So, $x - \frac{\pi}{3} = 2n\pi \pm \frac{\pi}{6}$. Step 4: Solve for $x$. Case 1: $x - \frac{\pi}{3} = 2n\pi + \frac{\pi}{6}$ $$ x = 2n\pi + \frac{\pi}{6} + \frac{\pi}{3} $$ $$ x = 2n\pi + \frac{\pi}{6} + \frac{2\pi}{6} $$ $$ x = 2n\pi + \frac{3\pi}{6} $$ $$ x = 2n\pi + \frac{\pi}{2} $$ Case 2: $x - \frac{\pi}{3} = 2n\pi - \frac{\pi}{6}$ $$ x = 2n\pi - \frac{\pi}{6} + \frac{\pi}{3} $$ $$ x = 2n\pi - \frac{\pi}{6} + \frac{2\pi}{6} $$ $$ x = 2n\pi + \frac{\pi}{6} $$ The general solution is $\boxed{x = 2n\pi + \frac{\pi}{2} \text{ or } x = 2n\pi + \frac{\pi}{6}, \text{ where } n \in \mathbb{Z}}$. That's 2 down. 3 left today — send the next one.