We are to find the maximum and the minimum values of the function $$f ( x ) = 4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x - 8 \sin 2 x - 7$$ where $0 \leqq x \leqq \pi$. Set $t = \sin x + \cos x$. Since $$\sin x + \cos x = \sqrt { \mathbf { A } } \sin \left( x + \frac { \mathbf { B } } { \mathbf { C } } \pi \right) , \quad ( \text { note: have } \mathbf { B } < \mathbf { C } )$$ the range of values which $t$ takes is $- \mathbf { D } \leqq t \leqq \sqrt { \mathbf{E} }$. Next, since $$\sin 2 x = t ^ { 2 } - \mathbf { F }$$ and $$4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x = - \mathbf { G } t ^ { 3 } + \mathbf { H } t ,$$ we have $$f ( x ) = - \mathbf { G } t ^ { 3 } - \mathbf { I } t ^ { 2 } + \mathbf { H } t + \mathbf { J } . \tag{1}$$ When we set the right side of (1) as $g ( t )$ and differentiate with respect to $t$, we have $$g ^ { \prime } ( t ) = - \mathbf { K } ( \mathbf { L } t - \mathbf { M } ) \left( t + \mathbf { N } \right) .$$ Hence at $t = \frac { \mathbf { O } } { \mathbf { P } } , g ( t ) ( = f ( x ) )$ takes the maximum value $\frac { \mathbf { Q R } } { \mathbf{S} }$, and at $t = \sqrt { \mathbf { U } }$, it takes the minimum value $\mathbf { V } \sqrt { \mathbf { W } } - \mathbf { X Y }$.
We are to find the maximum and the minimum values of the function
$$f ( x ) = 4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x - 8 \sin 2 x - 7$$
where $0 \leqq x \leqq \pi$.
Set $t = \sin x + \cos x$. Since
$$\sin x + \cos x = \sqrt { \mathbf { A } } \sin \left( x + \frac { \mathbf { B } } { \mathbf { C } } \pi \right) , \quad ( \text { note: have } \mathbf { B } < \mathbf { C } )$$
the range of values which $t$ takes is $- \mathbf { D } \leqq t \leqq \sqrt { \mathbf{E} }$.\\
Next, since
$$\sin 2 x = t ^ { 2 } - \mathbf { F }$$
and
$$4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x = - \mathbf { G } t ^ { 3 } + \mathbf { H } t ,$$
we have
$$f ( x ) = - \mathbf { G } t ^ { 3 } - \mathbf { I } t ^ { 2 } + \mathbf { H } t + \mathbf { J } . \tag{1}$$
When we set the right side of (1) as $g ( t )$ and differentiate with respect to $t$, we have
$$g ^ { \prime } ( t ) = - \mathbf { K } ( \mathbf { L } t - \mathbf { M } ) \left( t + \mathbf { N } \right) .$$
Hence at $t = \frac { \mathbf { O } } { \mathbf { P } } , g ( t ) ( = f ( x ) )$ takes the maximum value $\frac { \mathbf { Q R } } { \mathbf{S} }$, and at $t = \sqrt { \mathbf { U } }$, it takes the minimum value $\mathbf { V } \sqrt { \mathbf { W } } - \mathbf { X Y }$.