Consider the function
$$f ( x ) = \sin 2 x - 3 ( \sin x + \cos x )$$
on the interval $- \dfrac { \pi } { 3 } \leqq x \leqq \dfrac { \pi } { 3 }$.
(1) Let $t = \sin x + \cos x$. Find the range of the values which $t$ can take.
(2) The function $f ( x )$ takes its minimum value $\mathbf { E } - \mathbf { F } \sqrt{\mathbf{G}}$ at $x = \dfrac { \mathbf { H } } { \mathbf { I } }$.

Consider the following two equations in $x$
$$\sin 2x + a\cos x = 0 \tag{1}$$ $$\cos 2x + a\sin x = -2 \tag{2}$$
over the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$, where $a > 0$.
Let $a = \sqrt{2}$. Then the value of $x$ which satisfies (1) is
$$x = \frac{\mathbf{AB}}{\mathbf{A}}$$
However, at this $x$ the value of the left side of (2) is $\mathbf{DE}$, and so equation (2) does not hold. Hence, when $a = \sqrt{2}$, (1) and (2) have no common solution.
Now, let us find a value of $a$ such that (1) and (2) have a common solution, and also the common solution $x$.
First, from (1) we have
$$\sin x = \frac{\mathbf{FG}}{\mathbf{H}}a, \quad \cos 2x = \mathbf{I} - \frac{a^2}{\mathbf{J}}.$$
When we substitute these into (2), we obtain
$$a^2 = \mathbf{K}.$$
Thus $a = \sqrt{\mathbf{K}}$, and the common solution is
$$x = \frac{\mathbf{LM}}{\mathbf{N}}$$

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 $\mathbf { V } \sqrt { \mathbf { W } } - \mathbf { X Y }$.

Let $0 \leq \theta \leq 2 \pi$. All $\theta$ satisfying $\sin 2 \theta > \sin \theta$ and $\cos 2 \theta > \cos \theta$ can be expressed as $a \pi < \theta < b \pi$, where $a$ and $b$ are real numbers. What is the value of $b - a$?
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 3 } { 4 }$
(5) 1

$$3\sin x - 4\cos x = 0$$
Given this, what is the value of $|\cos 2x|$?
A) $\frac{3}{4}$
B) $\frac{3}{5}$
C) $\frac{4}{5}$
D) $\frac{7}{25}$
E) $\frac{9}{25}$

$$\cos x \cdot \cos 2x = \frac { 1 } { 16 \sin x }$$
Given this, what is $\sin 4x$?
A) $\frac { 1 } { 2 }$
B) $\frac { 2 } { 3 }$
C) $\frac { 1 } { 4 }$
D) $\frac { \sqrt { 2 } } { 2 }$
E) $\frac { \sqrt { 3 } } { 2 }$

$$\frac { \cot x } { \tan x + \cot x } = 4 \sin x - 3$$
Given this, what is the value of $\sin x$?
A) $3 - 2 \sqrt { 2 }$
B) $1 - \sqrt { 3 }$
C) $- 1 + \sqrt { 2 }$
D) $- 1 + \sqrt { 3 }$
E) $- 2 + 2 \sqrt { 2 }$

For $0 \leq x \leq 2 \pi$, $\cos \mathrm { x } + \sin 2 \mathrm { x } = \cot \mathrm { x }$ What is the sum of the $\mathbf { x }$ values that satisfy this equation?
A) $2 \pi$
B) $3 \pi$
C) $4 \pi$
D) $\frac { 5 \pi } { 2 }$
E) $\frac { 7 \pi } { 2 }$

For $0 \leq \mathrm { x } \leq \pi$,
$$\frac { \sin x \cdot \tan x } { 3 } = 1 - \cos x$$
What is the sum of the $\mathbf { x }$ values that satisfy this equation?
A) $\frac { \pi } { 3 }$
B) $\frac { 2 \pi } { 3 }$
C) $\frac { 4 \pi } { 3 }$
D) $\pi$
E) $2 \pi$

For $0 < x < \pi$,
$$\frac { \sin x \cdot \cos x } { \sin x + \cos x } = \frac { \sin x - \cos x } { 2 }$$
What is the sum of the $\mathbf { x }$ values that satisfy the equality?
A) $\frac { \pi } { 2 }$ B) $\frac { 5 \pi } { 4 }$ C) $\frac { 7 \pi } { 4 }$ D) $\pi$ E) $2 \pi$

Where $0 < x < \frac{\pi}{2}$,
$$\frac{1 + \tan x}{\cot x} \cdot \frac{\sin x - \cos x}{\sin x} = 2$$
if this holds, what is the value of $\sin x$?
A) $\frac{1}{3}$
B) $\frac{3}{5}$
C) $\frac{\sqrt{2}}{2}$
D) $\frac{\sqrt{3}}{2}$
E) $\frac{\sqrt{5}}{3}$