Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.

The double angle formulas yield
$$
a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x=\frac{\color{#C00000}{b\sin(2x) + (c-a^2)\cos(2x)}+(\color{#00A000}{c+a^2})}{2}\tag{1}
$$
Since the range of $a\cos(x)+b\sin(x)$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$, we get that the range of $(1)$ is
$$
\left[\frac{\color{#00A000}{c+a^2}+\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2},\dfrac{\color{#00A000}{c+a^2}-\color{#C00000}{\sqrt{(c-a^2)^2+b^2}}}{2}\right]\tag{2}
$$

Why the range of $\mathbf{a\cos(x)+b\sin(x)}$ is $\mathbf{\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]}$ :
The formula for rotating the point $(a,b)$ around the origin is
$$
(a,b)\stackrel{\text{rotate by }x}{\longrightarrow}(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))\tag{3}
$$
Thus, as $x$ varies from $0$ to $2\pi$, $(\color{#C00000}{a\cos(x)+b\sin(x)},b\cos(x)-a\sin(x))$ traces out a circle of radius $\color{#C00000}{\sqrt{a^2+b^2}}$. Thus, the first coordinate, $\color{#C00000}{a\cos(x)+b\sin(x)}$ varies between $-\color{#C00000}{\sqrt{a^2+b^2}}$ and $\color{#C00000}{\sqrt{a^2+b^2}}$.
