The value of $$\theta$$, when $$\sqrt3 \cos \theta + \sin \theta = 1 (0^\circ \leq \theta \leq 90^\circ)$$, is:

As per the given question,

$$\sqrt3 \cos \theta + \sin \theta = 1 (0^\circ \leq \theta \leq 90^\circ)$$

Dividing the 2 on both side of the equation,

$$\Rightarrow \dfrac{\sqrt3}{2}\times \cos \theta +\dfrac{1}{2}\times \sin \theta = \dfrac{1}{2}$$

Now, $$\cos 30 \cos \theta +\sin 30 \sin \theta = \cos 60$$

We know that $$\cos(\theta+\phi)=\cos \theta \cos \phi -\sin \phi \sin \theta$$

$$\Rightarrow \cos(\theta-30) = \cos 60$$

$$\Rightarrow \theta-30=60$$

$$\Rightarrow \theta=90^\circ$$

Hence $$\theta =90^\circ$$