Plain trigo

Relevant wiki: Trigonometric Equations - R method

The maximum value that the expression $\sin\theta + \sqrt{3}\cos\theta$ can give is $\sqrt{{1}^2 + {\sqrt{3}}^2}\\ = \sqrt{1+3}\\ = \sqrt{4}\\ = 2$ .

So, this expression has $\color{#3D99F6}{\boxed{\text{no}}}$ solution which yields $2.05 > 2$ as the answer.

$\begin{aligned} \sin \theta + \sqrt 3 \cos \theta & = 2.05 \\ 2 \left(\frac 12 \sin \theta + \frac {\sqrt 3}2 \cos \theta \right) & = 2.05 \\ 2 \left(\sin \theta \cos \frac \pi 3 + \sin \frac \pi 3 \cos \theta \right) & = 2.05 \\ 2 \sin \left(\theta + \frac \pi 3 \right) & = 2.05 \end{aligned}$

We note that the LHS has a maximum of 2, which is less than 2.05. Therefore, the LHS is always less than and never equal to the RHS and there is $\boxed{0}$ solution to the equation.

Since the range of $\theta$ is not stated, and the equation only has trigonometric functions, we know that the answer is either $0$ or $\infty$

$\sin \theta + \sqrt{3}\cos \theta = 2.05$

Let $f(x)=\sin \theta + \sqrt{3}\cos \theta$

$f'(x)=\cos \theta - \sqrt{3}\sin \theta$

The maximum value of $f(x)$ occurs when $f'(x)=0$ :

$\cos \theta - \sqrt{3}\sin \theta=0\\ \cos \theta = \sqrt{3}\sin \theta\\ \dfrac{\sin \theta}{\cos \theta} = \dfrac{1}{\sqrt{3}}\\ \tan \theta = \dfrac{1}{\sqrt{3}}\\ \theta = (30+180n)^{\circ}$

where $n$ is an integer.

If $|n|$ is even ( $\theta$ in quadrant I)

$f(x) = \sin(30+180n)^{\circ} + \sqrt{3}\cos(30+180n)^{\circ} \\ =\dfrac{1}{2}+\sqrt{3}\left(\dfrac{\sqrt{3}}{2}\right)\\ =\dfrac{1}{2}+\dfrac{3}{2}\\ =2$

If $|n|$ is odd ( $\theta$ in quadrant III)

$f(x) = \sin(30+180n)^{\circ} + \sqrt{3}\cos(30+180n)^{\circ} \\ =-\dfrac{1}{2}+\sqrt{3}\left(-\dfrac{\sqrt{3}}{2}\right)\\ =-\dfrac{1}{2}-\dfrac{3}{2}\\ =-2$

(Note: For a better proof, do the second derivative test)

The range of $f(x)$ is found to be $-2\leq f(x) \leq 2$ . From here, we can see that $f(x)=2.05$ is not achievable.

Therefore, there are $\boxed{0}$ solutions to this equation