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I have found two different methods of solving this trigonometric equation :

$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$ But these methods give different answers.

Solution 1

$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$

$$\implies \frac{1}{\text{cos}^2x} + \frac{3}{\text{sin}^2x} =8 $$ $$\implies 3{\text{cos}^2x} + {\text{sin}^2x} =8(\text{cos}^2x ×\text{sin}^2x)$$ $$\implies 2{\text{cos}^2x} + 1 =8(\text{cos}^2x ×(1-\text{cos}^2x))$$ $$\implies 8{\text{cos}^4x} -6\text{cos}^2x +1 =0$$ $$\implies (2\text{cos}^2x-1)(4\text{cos}^2x-1)=0 $$ $$ \implies x= 2nπ±\frac{π}{4},2nπ±(π-\frac{π}{4}),2nπ±\frac{π}{3},2nπ±(π-\frac{π}{3}), $$

Solution 2

$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$ $$ \implies 1+\text{tan}^2x + 3 + 3\text{cot}^2x =8 $$ $$\implies {\text{tan}^2x} + \frac{3}{\text{tan}^2x} =4 $$ $$ \implies \text{tan}^4x -4\text{tan}^2x + 3 =0 $$ $$ \implies (\text{tan}^2x -1)(\text{tan}^2x-3) =0 $$ $$ \implies x=nπ±\frac{π}{3},nπ±\frac{π}{4} $$

Now my question is that which one should I reject and why ?

ACB
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Regan Rogers
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  • All solutions are valid. I checked this on CAS too. Note that solution 2 encompasses solution 1. – Dstarred Dec 25 '22 at 07:11
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    You wrote $(4\cos^2 x-1) = 0$ and that seems to resolve to $|\cos x| = \frac 12$, which means that there should be $\frac{\pi}{3}$ appearing among those solutions. It's not there, which means that particular step is wrong. Instead, you seem to have written $\pi - \frac{\pi}{4}$ which is off. Also, both solutions need some more refinement at the last pair of lines. – Sarvesh Ravichandran Iyer Dec 25 '22 at 07:14
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    ....This would mean solution 2 states all possible solutions. Maybe recheck solution 1 again. – Dstarred Dec 25 '22 at 07:16
  • Sorry I forgot to add two other values of x in solution 1. Edited – Regan Rogers Dec 25 '22 at 07:19
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    @ReganRogers Welcome to Math SE. FYI, using an Approach0 search, there's How to solve $\sec^2x+3\csc^2x=8$, which provides several other ways to solve the problem (but this answer currently has a mistake, as the equation it derives of $8x^4+6x^2 - 1=0$ should be $8x^4-6x^2+1=0$ instead; note I've left a comment there to the author about this). – John Omielan Dec 25 '22 at 07:53
  • Both your solutions are the same. Note that the two branches $2n\pi\pm \frac{\pi}{3}$ and $2n\pi\pm (\pi-\frac{\pi}{3})$ in solution 1 can be merged under $n\pi\pm\frac{\pi}{3}$ in solution 2. Same thing happens with $2n\pi\pm \frac{\pi}{4}$ and $2n\pi\pm (\pi-\frac{\pi}{4})$ in solution 1 and $n\pi\pm\frac{\pi}{4}$ in solution 2. – Mostafa Ayaz Dec 25 '22 at 09:41

1 Answers1

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Using $\cos^{2}x + \sin^{2}x = 1$, from your first solution method and first factor, $2\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{2}$, so $\sin^{2}x = 1 - \frac{1}{2} = \frac{1}{2}$ and, thus, $\tan^{2}x = \frac{1/2}{1/2} = 1$. The second factor gives $4\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{4}$, so $\sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}$, which means $\tan^{2}x = \frac{3/4}{1/4} = 3$. Note these match the $\tan^{2}x$ values obtained from your second solution method's factorization.

Your first two sets of values of $x= 2n\pi\pm\frac{\pi}{4},2n\pi\pm(\pi-\frac{\pi}{4})$ can be combined into just $n\pi\pm\frac{\pi}{4}$ (since the $+$ part of $2n\pi\pm(\pi-\frac{\pi}{4})$ is $(2n+1)\pi-\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi-\frac{\pi}{4}$, with the $-$ part being $(2n-1)\pi+\frac{\pi}{4}$, i.e., all odd values of $n$ with $n\pi+\frac{\pi}{4}$), with this just being the solutions to $\cos^{2}x = \frac{1}{2} \; \to \; \cos x = \pm\frac{1}{\sqrt{2}}$. For the second set of solutions, $\cos^{2}x = \frac{1}{4} \; \to \; \cos x = \pm\frac{1}{2}$, which has solutions of $n\pi\pm\frac{\pi}{3}$ (your second set of values have the same basic issue as with the first set, i.e., they can be combined into just the one set of values I give here, and which you also gave in your second solution).

Thus, the sets of $x$ values from each of your $2$ solution techniques actually match each other.

John Omielan
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