Derivatives Of Functions In Parametric Forms

To solve the given equation and find the least value of such that this equation has exactly 7 solutions in the interval , start by analyzing the trigonometric identity:

We know that and . Therefore, the equation can be expressed as:

Simplifying gives:

Let , making the equation:

Solving this quadratic equation using the quadratic formula gives:

Plugging in the values where , , :

This results in two potential solutions for :

1. \)
2. \)

Since , for real angles , only is valid (as should be positive and refers to a value greater than or equal to 1 or less than or equal to -1).

For \) , we have:

\)

The general solution for within cycle of to is:

, where is an integer.

The function will repeat for every interval.

Now, find the smallest such that there are exactly 7 solutions to in the interval .

Each period from contains 2 solutions. We need at least 4 periods such that:

The interval length (for 4 full cycles, as every cycle causes 2 solutions).

Thus, implies:

Finally, calculate:

Using the formula for such series:

This implies:

So, the final answer matches the option: