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.
$$\sec^2x+3\csc^2x=8$$
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1start with multiplying both sides by $\sin^2 x \cos^2 x$. – Math-fun Mar 07 '15 at 16:02
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@grargside: Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – AlexR Mar 07 '15 at 16:05
3 Answers
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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 $\cos2x=\cos2A\implies2x=2m\pi\pm2A$
lab bhattacharjee
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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 $$x^2 = \frac{-6 \pm\sqrt{68}}{16}, \, x = \pm\sqrt{\frac{-6 +\sqrt{68}}{16}}, t = 81.92^\circ, 98.07^\circ, 261.92^\circ, 278.07^\circ $$
abel
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Note your $8x^4+6x^2 - 1=0$ should be $8x^4-6x^2+1=0$ instead, with this fairly easily factoring as $(4x^2-1)(2x^2-1)=0$. – John Omielan Dec 25 '22 at 07:54
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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
lab bhattacharjee
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