If $$3 \cos^2 A + 6 \sin^2 A = 3, 0^\circ \leq A \leq 90^\circ$$, then the value of A is:

It is given that,

$$3 \cos^2 A + 6 \sin^2 A = 3------(i)$$

And  $$0^\circ \leq A \leq 90^\circ$$

Now, We can write the equation as below

$$3 \cos^2 A + 3 \sin^2 A+3\sin^2 A = 3$$
$$\Rightarrow 3 (\cos^2 A + \sin^2 A)+3\sin^2 A = 3$$
From the trigonometric identity, we know that $$\cos^2 A + \sin^2 A=1$$
$$\Rightarrow 3 \times 1+3\sin^2 A = 3$$
$$\Rightarrow 3 +3\sin^2 A = 3$$
$$\Rightarrow 3\sin^2 A = 0$$
$$\Rightarrow \sin A = 0=\sin 0$$
Hence, $$A=0$$