# $\text{sec}^2x + 3\text{cosec}^2x =8$ Solving this trigonometric equation.

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 ?

• All solutions are valid. I checked this on CAS too. Note that solution 2 encompasses solution 1. Commented Dec 25, 2022 at 7:11
• 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. Commented Dec 25, 2022 at 7:14
• ....This would mean solution 2 states all possible solutions. Maybe recheck solution 1 again. Commented Dec 25, 2022 at 7:16
• Sorry I forgot to add two other values of x in solution 1. Edited Commented Dec 25, 2022 at 7:19
• @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). Commented Dec 25, 2022 at 7:53

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.