Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression

$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$

My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\theta \\
&=3(1-2\sin\theta \cos\theta)+2\sin^2\theta \\
&=3(\sin\theta-\cos\theta)^2 + 2\sin^2\theta
\end{align}$$
Hence the minimum value of $f(\theta)=2\sin^2\theta$ when $\theta=\pi/4$
hence minimum value of $f(\theta)=1$.
But then again I tried to do question differently by making substitutions in order to change the whole $f(\theta)$ in the form of $\cos x+\sin x$ 
Then $f(\theta)$ came out to be 
$$f(\theta)= 4-(\cos(2\theta)+3\sin(2\theta))$$
The minimum value of this expression is surely $$(f(\theta))_{min}=4-\sqrt{10}$$
Can anybody explain me algebraically why my first method gave the wrong result?

$f(\theta) = 4 - \sqrt{10}$ is correct.
so what is the error here:
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + 2\sin^2\theta$
That part is a true statement but then you say.
Hence the minimum value of $f(θ)=2\sin^2θ,$ when $θ=π/4$ hence minimum value of $f(θ)=1.$
It is a logical jump that was a step too far.
If you had had, 
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + k$
it would be okay to zero out the term in parentheses.  It must be greater than or equal to zero, so set it to zero.
But in your actual expression
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + 2\sin^2\theta$
Minimizing either term doesn't minimize the sum.
Furthermore, when you say the minimum of $f(θ)=2\sin^2θ,$ occurs when  $θ=\pi/4$ that is just wrong.
