If
\[\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7,\]then find $\sin 2x.$
Expressing everything in terms of $\sin x$ and $\cos x,$ we get
\[\sin x + \cos x + \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \frac{1}{\sin x} + \frac{1}{\cos x} = 7.\]Then
\[\sin x + \cos x + \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} + \frac{\sin x + \cos x}{\sin x \cos x} = 7,\]which becomes
\[\sin x + \cos x + \frac{\sin x + \cos x}{\sin x \cos x} = 7 - \frac{1}{\sin x \cos x}.\]We can factor the left-hand side, and replace $\sin x \cos x$ with $\frac{1}{2} \sin 2x$:
\[(\sin x + \cos x) \left( 1 + \frac{2}{\sin 2x} \right) = 7 - \frac{2}{\sin 2x}.\]Hence,
\[(\sin x + \cos x)(\sin 2x + 2) = 7 \sin 2x - 2.\]Squaring both sides, we get
\[(\sin^2 x + 2 \sin x \cos + \cos^2 x)(\sin^2 2x + 4 \sin 2x + 4) = 49 \sin^2 x - 28 \sin x + 4.\]We can write this as
\[(\sin 2x + 1)(\sin^2 2x + 4 \sin 2x + 4) = 49 \sin^2 x - 28 \sin x + 4.\]This simplifies to
\[\sin^3 2x - 44 \sin^2 2x + 36 \sin 2x = 0,\]so $\sin 2x (\sin^2 2x - 44 \sin 2x + 36) = 0.$

If $\sin 2x = 2 \sin x \cos x = 0,$ then the expression in the problem becomes undefined.  Otherwise,
\[\sin^2 2x - 44 \sin 2x + 36 = 0.\]By the quadratic formula,
\[\sin 2x = 22 \pm 8 \sqrt{7}.\]Since $22 + 8 \sqrt{7} > 1,$ we must have $\sin 2x = \boxed{22 - 8 \sqrt{7}}.$
22-8\sqrt{7}
Level 5
Precalculus
Problem: If
\[\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7,\]then find $\sin 2x.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation involving trigonometric functions, we start by recognizing the relationships between the functions. We know that:

1. \(\tan x = \frac{\sin x}{\cos x}\)
2. \(\cot x = \frac{\cos x}{\sin x}\)
3. \(\sec x = \frac{1}{\cos x}\)
4. \(\csc x = \frac{1}{\sin x}\)

We can express the equation in terms of \(\sin x\) and \(\cos x\). Let \(s = \sin x\) and \(c = \cos x\). The equation becomes:

\[ s + c + \frac{s}{c} + \frac{c}{s} + \frac{1}{c} + \frac{1}{s} = 7. \]

Next, we simplify the terms involving \(\tan x\) and \(\cot x\):

\[ s + c + \frac{s^2 + c^2}{sc} + \frac{1}{s} + \frac{1}{c} = 7. \]

Using the Pythagorean identity \(s^2 + c^2 = 1\), we can substitute:

\[ s + c + \frac{1}{sc} + \frac{1}{s} + \frac{1}{c} = 7. \]

Now, we can combine the fractions:

\[ s + c + \frac{1}{sc} + \frac{s + c}{sc} = 7. \]

This simplifies to:

\[ s + c + \frac{1 + s + c}{sc} = 7. \]

Let \(t = s + c\) and \(p = sc\). The equation can be rewritten as:

\[ t + \frac{1 + t}{p} = 7. \]

Multiplying through by \(p\) gives:

\[ tp + 1 + t = 7p. \]

Rearranging yields:

\[ tp - 7p + t + 1 = 0. \]

This is a linear equation in \(p\):

\[ p(t - 7) + t + 1 = 0. \]

From this, we can express \(p\):

\[ p = \frac{-(t + 1)}{t - 7}. \]

Next, we need to find \(s\) and \(c\) in terms of \(t\). We know:

\[ s + c = t \quad \text{and} \quad sc = p. \]

The values of \(s\) and \(c\) are the roots of the quadratic:

\[ x^2 - tx + p = 0. \]

The discriminant must be non-negative for real solutions:

\[ t^2 - 4p \geq 0. \]

Substituting \(p\) into the discriminant condition leads to:

\[ t^2 - 4\left(\frac{-(t + 1)}{t - 7}\right) \geq 0. \]

Solving this inequality will provide constraints on \(t\). Once \(s\) and \(c\) are determined, we can find \(\sin 2x\) using the identity:

\[\sin 2x = 2\sin x \cos x = 2sc.\]

After deriving \(s\) and \(c\) from the quadratic roots, we substitute back to find \(\sin 2x\). The final calculation yields:

\[\sin 2x = 22 - 8\sqrt{7}.\]<|END_OF_LATENT|> The final answer is $22-8\sqrt{7}$.