Explanation
$$\sin x + 2 \sin2x - \sin 3 x = 3$$
$$\sin x+4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 3$$
$$\sin x [-2 + 4 \cos x + 4 (1 - \cos^2 x)] =3$$
$$\sin x [2 - (4 \cos^2 x - 4 \cos x +1)+1] =3$$
$$\sin x [3 - (2 \cos x - 1)^2] =3$$
$$\Rightarrow \sin x = 1$$ and $$2 \cos x - 1 =0$$
$$\Rightarrow x = \dfrac{\pi}{2}$$ and $$x = \dfrac{\pi}{3}$$
Which is not possible at same time.
Hence, no solution.
Consider the given expression,
$${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha $$
AS it is clear that $${{\cos }^{2}}\alpha $$ and $${{\sec }^{2}}\alpha $$ are positive numbers
So, we can apply the theorem of A.M and G.M
A.M>=G.M
so we have
$$\left( \dfrac{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }{2} \right)=\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }$$
Or $$ {{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2\sqrt{{{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha }=2.1 $$
$$ {{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha =2 $$
So, minimum value of $${{\sec }^{2}}\alpha +{{\cos }^{2}}\alpha $$ is 2.
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