Question

3sec^4 x - 10sec^2 x - 8 = 0, what is sin x

Answer 1

Set sec^2 as y, solve quadratic for sec2, sub for 1 + tan^2 etc...

Answer 2

\[(3\sec^{2}(x) - 4)(\sec^{2}(x) - 2) = 0\]
\[\sec^{2}(x) = \frac{4}{3} \space\vee\space \sec^{2}(x) = 2\]
\[\cos^2(x) = \frac{3}{4} \space\vee\space \cos^2(x) = \frac{1}{2}\]
\[1-\sin^{2}(x) = \frac{3}{4} \space\vee\space 1-\sin^{2}(x) = \frac{1}{2}\]
\[\sin(x) = \sqrt{\frac{1}{4}} \space\vee\space \sin(x) = \sqrt{\frac{1}{2}}\]

Answer 3

I'm very sorry, I made a very stupid mistake:
\[(3\sec^2(x) + 2)(\sec^2(x) - 4) = 0\]
\[\sec^2(x) = -\frac{2}{3} \space\vee\space \sec^2(x) = 4\]
\[\cos^2(x) = -\frac{3}{2} \space\vee\space \cos^2(x) = \frac{1}{4}\]
\[\sin^2(x) = \frac{5}{2} \space\vee\space \sin^2(x) = \frac{3}{4}\]
There is no real inverse sine of 5/2, so only carry on with the latter:
\[\sin(x) = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}\]
Please accept my apologies for the incorrect answer.