How to solve $\sec^2x+3\csc^2x=8$ I have tried substitution involving the Pythagorean Identities, as well as converting to just sin and cos.  I cannot seem to manipulate the equation into a form which has a clear solution process.  Please provide help.
$$\sec^2x+3\csc^2x=8$$
 A: HINT:
Write $\sec^2x=1+\tan^2x, \csc^2x=1+\cot^2x=1+\dfrac1{\tan^2x}$ to form a Quadratic eqaution in $\tan^2x$
Then use  $\tan^2x=\tan^2A\implies x=n\pi\pm A$ where $n$ is any integer
Or use $\cos2x=\dfrac{1-\tan^2x}{1+\tan^2x}$ and $\cos2x=\cos2A\implies2x=2m\pi\pm2A$ 
A: it is some time easier to the unit circle. we will take $$x = \cos t, y = \sin t $$ the important relation to remember is  $$x^2 + y^2 = 1. \tag 1$$ we can convert  $$\sec^2 t + 3\csc^2 t = 8 $$ into $$\frac 1{x^2} + \frac{3}{y^2} = 8$$ so you have $$3x^2 + y^2 = 8x^2 y^2 \tag 2 $$
you can solve $(1)$ and $(2)$ by eliminating ,say, $y$ to get $$3x^2 + 1 - x^2 =8x^2(1-x^2)$$ this is a quadratic equation in $x^2.$  that is $$8x^4+6x^2 - 1=0$$ has roots $$x^2 = \frac{-6 \pm\sqrt{68}}{16}, \, x = \pm\sqrt{\frac{-6 +\sqrt{68}}{16}}, t = 81.92^\circ, 98.07^\circ, 261.92^\circ, 278.07^\circ  $$
A: HINT:
$$\sec^2x=\frac1{\cos^2x}\text{ and }\csc^2x=\frac1{\sin^2x}$$
$$\cos2x=1-2\sin^2x=2\cos^2x-1$$ to form a Quadratic Eqaution in $\cos2x$ on simplification 
