If $$2 \tan ^2 \theta-5 \sec \theta=1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n \pi}{2}\right]$$, for the least value of $$n \in \mathbf{N}$$, then $$\sum_\limits{k=1}^n \frac{k}{2^k}$$ is equal to:

The number of elements in the set

$$S=\left\{\theta \in[0,2 \pi]: 3 \cos ^{4} \theta-5 \cos ^{2} \theta-2 \sin ^{6} \theta+2=0\right\}$$ is :

Let $$S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}$$ and $$\beta=\sum_\limits{x \in S} \tan ^{2}\left(\frac{x}{3}\right)$$, then $$\frac{1}{6}(\beta-14)^{2}$$ is equal to :

The number of elements in the set $$S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$$ is :