1. **Problem Statement:** We have $$f(x) = \sqrt{3}\cos x + \sin x$$ Express $f(x)$ in the form $$R \cos(x - \lambda)$$ where $R > 0$ and $0 < \lambda < 90^\circ$. Then find the general solution of $$f(x) = \sqrt{3}$$ and find the minimum and maximum values of $$\frac{1}{5} + f(x)$$. 2. **Expressing $f(x)$ in the form $R \cos(x - \lambda)$:** We use the identity $$a \cos x + b \sin x = R \cos(x - \lambda)$$ where $$R = \sqrt{a^2 + b^2}$$ and $$\tan \lambda = \frac{b}{a}$$ with $R > 0$ and $0 < \lambda < 90^\circ$. 3. Identify coefficients: $$a = \sqrt{3}, \quad b = 1$$ Calculate $R$: $$R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$ Calculate $\lambda$: $$\tan \lambda = \frac{1}{\sqrt{3}}$$ Since $0 < \lambda < 90^\circ$, then $$\lambda = 30^\circ = \frac{\pi}{6}$$ 4. Rewrite $f(x)$: $$f(x) = 2 \cos\left(x - \frac{\pi}{6}\right)$$ 5. **Find the general solution of** $$f(x) = \sqrt{3}$$ which is $$2 \cos\left(x - \frac{\pi}{6}\right) = \sqrt{3}$$ Divide both sides by 2: $$\cos\left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$ The general solutions for $\cos \theta = \frac{\sqrt{3}}{2}$ are $$\theta = \pm \frac{\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ So: $$x - \frac{\pi}{6} = \pm \frac{\pi}{6} + 2k\pi$$ Case 1: $$x - \frac{\pi}{6} = \frac{\pi}{6} + 2k\pi \Rightarrow x = \frac{\pi}{3} + 2k\pi$$ Case 2: $$x - \frac{\pi}{6} = -\frac{\pi}{6} + 2k\pi \Rightarrow x = 2k\pi$$ Thus the general solution is $$x = 2k\pi \quad \text{or} \quad x = \frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ 6. **Find minimum and maximum values of** $$\frac{1}{5} + f(x) = \frac{1}{5} + 2 \cos\left(x - \frac{\pi}{6}\right)$$ Since $$\cos\left(x - \frac{\pi}{6}\right)$$ has minimum $-1$ and maximum $1$, then: Minimum: $$\frac{1}{5} + 2(-1) = \frac{1}{5} - 2 = -\frac{9}{5} = -1.8$$ Maximum: $$\frac{1}{5} + 2(1) = \frac{1}{5} + 2 = \frac{11}{5} = 2.2$$ **Final answers:** - $f(x) = 2 \cos\left(x - \frac{\pi}{6}\right)$ with $R=2$, $\lambda=\frac{\pi}{6}$ (30 degrees) - General solution of $f(x) = \sqrt{3}$ is $$x = 2k\pi \quad \text{or} \quad x = \frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ - Minimum value of $\frac{1}{5} + f(x)$ is $-1.8$ - Maximum value of $\frac{1}{5} + f(x)$ is $2.2$