Hello again jhunt47

This is an interesting one! Originally Posted by

**jhunt47** Solve the Following Expression:

(16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!

First notice that$\displaystyle \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot\left(\fra c{16}{81}\right)^{-\sin^2x}$

So, if we let $\displaystyle y = \left(\frac{16}{81}\right)^{\sin^2x}$, then$\displaystyle \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot y^{-1}=\left(\frac{16}{81}\right)\cdot\frac{1}{y}=\fra c{16}{81y}$

and the equation becomes$\displaystyle y+\frac{16}{81y}=\frac{26}{27}$

$\displaystyle \Rightarrow 81y^2+16=78y$

$\displaystyle \Rightarrow 81y^2-78y+16=0$

$\displaystyle \Rightarrow (3y-2)(27y-8)=0$

$\displaystyle \Rightarrow y = \left(\frac{16}{81}\right)^{\sin^2x}= \frac23, \,\frac{8}{27}$

We now note that $\displaystyle \left(\frac{16}{81}\right)^{\frac14}=\frac23$ and $\displaystyle \left(\frac{16}{81}\right)^{\frac34}=\frac{3}{27}$$\displaystyle \Rightarrow\sin^2x=\frac14,\,\frac34$

$\displaystyle \Rightarrow \sin x = \pm\frac12,\,\pm\frac{\sqrt3}{2}$

$\displaystyle \Rightarrow x = n\pi \pm \frac{\pi}{6}$ or $\displaystyle n\pi \pm \frac{\pi}{3},\;n \in \mathbb{Z}$

Grandad