find the min of

There are many approaches to this problem, as such i will share 2 approaches.

Cauchy-Swartz: $\left(1+\dfrac{2}{3}\right)\left(\sin^4 (x)+\dfrac{3}{2} \cos^4(x)\right)\geq \left(\sin^2 (x) +\cos^2(x)\right)^2 \to \sin^4 (x)+\dfrac{3}{2} \cos^4(x)\geq \boxed{\dfrac{3}{5}}$ the equality case being when $\dfrac{\sin^4(x)}{1} =\dfrac{ \dfrac{3}{2} \cos^4(x)}{2/3}\to \sin^2 (x) = \dfrac{3}{5}$

quadratic

we can write this as $\left(\sin^2(x)\right)^2 +\dfrac{3}{2} \left(1-\sin^2(x)\right)^2 = \dfrac{5}{2} \left(\sin^2(x)-\dfrac{3}{5} \right)^2+\boxed{\dfrac{3}{5}}$ and this is minimized when $\sin^2(x)-\dfrac{3}{5}=0$

We have $f\left(x\right) = \sin ^{4} \left( x \right) + \frac{3}{2}\cos ^{4} \left( x \right)$ we take the derivative which yields $f'\left(x\right) = 4\sin ^{3} \left( x \right) \cos \left( x \right) + 6\cos ^{3} \left( x \right) \left( - \sin \left( x \right) \right)$ Though we can factor right hand side so we have $\sin \left( x \right) \cos \left( x \right) \left( 4\sin ^2 \left( x \right) - 6\cos ^2 \left( x \right) \right)$ But we can factor a 2 out to get $2\sin \left( x \right) \cos \left( x \right) \left( 2\sin ^2 \left( x \right) - 3\cos ^2 \left( x \right) \right)$ recall the identity $\sin \left( 2x \right) = 2\sin \left( x \right) \cos \left( x \right)$ and $\sin ^2 \left( x \right) + \cos ^2 \left( x \right) = 1\Leftrightarrow - 3\cos \left( x \right) = 3\sin ^2 \left( x \right) - 3$ so we have $f'\left(x\right) = \sin \left( 2x \right) \left( 5\sin ^2 \left( x \right) -3 \right)$ solutions are when $x= 0\text{ and } \sin ^2 \left( x \right) = \frac{3}{5}$ . We check whether they are minimum or maximum points by analyzing the second derivative. We have $f''\left(x\right) = \cos \left( 2x \right) 2\left( 5\sin ^2 \left( x \right) -3 \right) + \sin \left( 2x \right) \left( 10\sin \left( x \right) \cos \left( x \right) \right)$ We have $f''\left(0\right)= -6$ which means this is a maximum thus the minimum must be at $\sin ^2 \left( x \right) = \frac{3}{5}$ , thus we go back to the original equation which is $f\left(x\right) = \left( \sin ^2 \left( x \right) \right) ^{2} + \frac{3}{2}\left( 1 - \sin ^2 \left( x \right) \right) ^{2}$ evaluating when $\sin ^2 \left( x \right) = \frac{3}{5}$ we get $\left( \frac{3}{5} \right) ^2 + \frac{3}{2}\left( 1 - \frac{3}{5} \right) ^2 = \frac{9}{25} + \frac{6}{25}= \frac{15}{25}= \frac{60}{100}= \boxed{0.6}$

$\frac{\partial \left(\sin ^4(x)+\frac{3 \cos ^4(x)}{2}\right)}{\partial x} \Rightarrow 4 \sin ^3(x) \cos (x)-6 \sin (x) \cos ^3(x)$

Solve $4 \sin ^3(x) \cos (x)-6 \sin (x) \cos ^3(x)=0$ for the principal values (they repeat every $2\pi$ ): $\begin{array}{l} x\to 0 \\ x\to -\frac{\pi }{2} \\ x\to \frac{\pi }{2} \\ x\to \pi \\ x\to -\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right) \\ x\to \pi -\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right) \\ x\to \tan ^{-1}\left(\sqrt{\frac{3}{2}}\right) \\ x\to \tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)-\pi \\ \end{array}$

Evaluating $\sin ^4(x)+\frac{3 \cos ^4(x)}{2}$ at those values of $x$ gives: $\frac{3}{2},1,1,\frac{3}{2},\frac{3}{5},\frac{3}{5},\frac{3}{5},\frac{3}{5}$

This are the minima, maxima and saddle points.

$\frac{\partial \frac{\partial \left(\sin ^4(x)+\frac{3 \cos ^4(x)}{2}\right)}{\partial x}}{\partial x} \Rightarrow -4 \sin ^4(x)-6 \cos ^4(x)+30 \sin ^2(x) \cos ^2(x)$

Evaluating $\text{sgn}\left(-4 \sin ^4(x)-6 \cos ^4(x)+30 \sin ^2(x) \cos ^2(x)\right)$ gives $\{-1,-1,-1,-1,1,1,1,1\}$ .

The $-1$ values are maxima. The $1$ values are minima. If $0$ values had occurred, then those would have to be checked further and probably would be saddle points.

The minimum is $\frac35$ .