21given = cos2x(tan2x+ 3tanx+ 5)
= (tan2x+ 3tanx+ 5)1+tan2x
= 1+ 3tanx + 41+tan2x
let tanx = m (any real number)
it bcomes 1+3m+41+m2
I have used calculus to get min{3m+41+m2} = -12 (when m = -3)
Hence minimum of given expression is 1+ (-12)= 12 which is obtained at x = tan-1(-3)
1708\hspace{-16}$Let $\mathbf{y=\sin^2 x+3\sin x.\cos x+5\cos^2 x}$\\\\ $\mathbf{y=\frac{1}{2}.\left(2\sin^2x+6\sin x.\cos x+10\cos^2 x\right)}$\\\\ $\mathbf{y=\frac{1}{2}.\left\{\sin^2 x+6\sin x.\cos x+(3\cos x)^2+\sin^2 x+\cos^2 x\right\}}$\\\\ $\mathbf{y=\frac{1}{2}.\left\{\left(\sin x+3.\cos x\right)^2+1\right\}\geq \frac{1}{2}}$\\\\ So $\boxed{\boxed{\mathbf{y_{Min.}=\frac{1}{2}}}}$ at $\mathbf{\tan x=-3}$
11Also, if the given exp. is denoted by M, then
2M=2\sin^2x+3\sin2x+10\cos^2x=8\cos^2x+2+3\sin2x
That gives 2M=4(2\cos^2x-1)+6+3\sin2x=4\cos2x+3\sin2x+6\ge1
Implies M\ge \frac{1}{2}
341sin2x + 3 sin x cos x + 5 cos2x = 1+2(1+cos 2x)+3/2 sin 2x
= 3 + 2 cos 2x + 3/2 sin 2x
Now we know that the min of a sin x + b cos x +c is c-√a2+b2
Here it evaluates to 3 - 5/2 = 1/2.
Edit: oops, Soumik had done the same thing
