Question

Solve ${16}^{si{n}^{2}x}+{16}^{co{s}^{2}x}=10$

Answer 1

${16}^{si{n}^{2}x}+{16}^{co{s}^{2}x}=10$
or ${16}^{si{n}^{2}x}+{16}^{1-si{n}^{2}x}=10$
or ${16}^{si{n}^{2}x}+{16.16}^{-si{n}^{2}x}=10$
put ${16}^{si{n}^{2}x}=t$
Then $t+\frac{16}{t}=10or{t}^2-10t+16=0$
or $(t-2)(t-8)=0$ , giving $t=2or8$
Hence ${16}^{si{n}^{2}x}=2$ or $2^{4si{n}^{2}x}=2$, which gives
$4si{n}^{2}x=1$
or $si{n}^{2}x=\frac{1}{4}=si{n}^2\left(\frac{\pi }{6}\right)$.....(1)
$\therefore$ solution of (1) is, $x=n\pi \pm \frac{\pi }{6}$
where $n=0,\pm 1,\pm 2,\pm 3$ , .......
Again $t=8gives{2}^{4si{n}^{2}x}=8=2^3$
or $4si{n}^{2}x=3i.e.si{n}^{2}x=\frac{3}{4}=si{n}^2\frac{\pi }{3}$....(2)
$\therefore$ The solution of (2) is , $x=n\pi \pm \frac{\pi }{3}$
where $n=0,\pm 1,\pm 2,\pm 3$.......