Let S = x (-2,2) : 9^1-^2x+9^^2x=10 and b = _x S^2( x3).

Question

Let $S = \left\{x \in \left(-\dfrac\pi2,\dfrac\pi2ight) : 9^{1-	an^2x}+9^{	an^2x}=10ight\}$ and $b = \displaystyle\sum_{x\in S}	an^2\left(\dfrac x3ight)$. Then $\dfrac16(b-14)^2$ is equal to:

Accepted Answer

Let $P = 9^{\tan^2x}$. Then $\dfrac9P+P=10 \implies P^2-10P+9=0 \implies (P-9)(P-1)=0 \implies P=1,9$.

So $\tan^2x = 0$ or $1$, giving $x = 0,\ \dfrac\pi4,\ -\dfrac\pi4$ within $\left(-\dfrac\pi2,\dfrac\pi2\right)$, i.e. $S=\left\{0,\tfrac\pi4,-\tfrac\pi4\right\}$.

$$b = \tan^2(0) + \tan^2\left(\frac\pi{12}\right) + \tan^2\left(-\frac\pi{12}\right) = 2\tan^2(15^\circ) = 2(2-\sqrt3)^2 = 2(7-4\sqrt3) = 14-8\sqrt3$$

$$\frac16(b-14)^2 = \frac16(-8\sqrt3)^2 = \frac16(192) = 32$$

Solutions · 1