Let `f(x)= cos x sin 2x ` then choose the best option :

Options

A. `min underset(x in (-pi,pi))(f(x)) gt -7/9`
B. `min underset(x in (-pi,pi))(f(x)) gt -9/7`
C. `min underset(x in (-pi,pi))(f(x)) gt -1/9`
D. `min underset(x in (-pi,pi))(f(x)) gt -2/9`

Answer

To solve the problem, we need to analyze the function \( f(x) = \cos x \sin 2x \) and find its minimum value over the interval \( x \in [-\pi, \pi] \).

### Step 1: Rewrite the function
We can express \( \sin 2x \) using the double angle identity:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite \( f(x) \):
\[ f(x) = \cos x \cdot (2 \sin x \cos x) = 2 \sin x \cos^2 x \]

### Step 2: Substitute \( t = \sin x \)
Let \( t = \sin x \). The range of \( t \) when \( x \) varies from \( -\pi \) to \( \pi \) is \( t \in [-1, 1] \). We can express \( \cos^2 x \) in terms of \( t \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - t^2 \]
Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = 2t(1 - t^2) = 2t - 2t^3 \]

### Step 3: Define the new function \( g(t) \)
Let \( g(t) = 2t - 2t^3 \). We will find the minimum of \( g(t) \) over the interval \( t \in [-1, 1] \).

### Step 4: Differentiate \( g(t) \)
To find the critical points, we differentiate \( g(t) \):
\[ g'(t) = 2 - 6t^2 \]
Setting the derivative equal to zero to find critical points:
\[ 2 - 6t^2 = 0 \implies 6t^2 = 2 \implies t^2 = \frac{1}{3} \implies t = \pm \frac{1}{\sqrt{3}} \]

### Step 5: Evaluate the second derivative
To determine whether these critical points are minima or maxima, we compute the second derivative:
\[ g''(t) = -12t \]
- At \( t = \frac{1}{\sqrt{3}} \):
\[ g''\left(\frac{1}{\sqrt{3}}\right) = -12 \cdot \frac{1}{\sqrt{3}} < 0 \quad \text{(local maximum)} \]
- At \( t = -\frac{1}{\sqrt{3}} \):
\[ g''\left(-\frac{1}{\sqrt{3}}\right) = 12 \cdot \frac{1}{\sqrt{3}} > 0 \quad \text{(local minimum)} \]

### Step 6: Evaluate \( g(t) \) at critical points and endpoints
Now we evaluate \( g(t) \) at the critical points and the endpoints \( t = -1 \) and \( t = 1 \):
- At \( t = -1 \):
\[ g(-1) = 2(-1) - 2(-1)^3 = -2 + 2 = 0 \]
- At \( t = 1 \):
\[ g(1) = 2(1) - 2(1)^3 = 2 - 2 = 0 \]
- At \( t = \frac{1}{\sqrt{3}} \):
\[ g\left(\frac{1}{\sqrt{3}}\right) = 2\left(\frac{1}{\sqrt{3}}\right) - 2\left(\frac{1}{\sqrt{3}}\right)^3 = \frac{2}{\sqrt{3}} - 2\left(\frac{1}{3\sqrt{3}}\right) = \frac{2}{\sqrt{3}} - \frac{2}{3\sqrt{3}} = \frac{6 - 2}{3\sqrt{3}} = \frac{4}{3\sqrt{3}} \]

### Step 7: Compare values
Now we compare:
- \( g(-1) = 0 \)
- \( g(1) = 0 \)
- \( g\left(-\frac{1}{\sqrt{3}}\right) \) (local minimum)

To find \( g\left(-\frac{1}{\sqrt{3}}\right) \):
\[ g\left(-\frac{1}{\sqrt{3}}\right) = 2\left(-\frac{1}{\sqrt{3}}\right) - 2\left(-\frac{1}{\sqrt{3}}\right)^3 = -\frac{2}{\sqrt{3}} + \frac{2}{3\sqrt{3}} = -\frac{6 - 2}{3\sqrt{3}} = -\frac{4}{3\sqrt{3}} \]

### Conclusion
The minimum value of \( f(x) \) occurs at \( t = -\frac{1}{\sqrt{3}} \) and is:
\[ \text{Minimum of } f(x) = -\frac{4}{3\sqrt{3}} \approx -0.77 \]
This value is greater than \( -\frac{7}{9} \), confirming that \( f(x) > -\frac{7}{9} \) for \( x \in [-\pi, \pi] \).

### Final Answer
The best option is \( \text{Option E} \).