How do you solve #1+tan^2x=6-2sec^2x#?

Question

Anna Andrews · Accepted Answer

I found 4 possible values for #x# between #0 and 2pi#:

Have a look:

Repeating at each #2kpi#

Anna Andrews · Answer

undefined To solve $1 + \tan^2(x) = 6 - 2\sec^2(x)$, first, express $\tan^2(x)$ and $\sec^2(x)$ in terms of $\sin(x)$ and $\cos(x)$:

$$1 + \frac{\sin^2(x)}{\cos^2(x)} = 6 - 2\frac{1}{\cos^2(x)}$$

Now, bring all terms to one side of the equation:

$$1 + \frac{\sin^2(x)}{\cos^2(x)} - 6 + 2\frac{1}{\cos^2(x)} = 0$$

$$2\frac{\sin^2(x)}{\cos^2(x)} - 2\frac{1}{\cos^2(x)} - 5 = 0$$

$$2\frac{\sin^2(x) - 1}{\cos^2(x)} - 5 = 0$$

$$\frac{2(\sin^2(x) - 1)}{\cos^2(x)} - 5 = 0$$

$$\frac{2\sin^2(x) - 2}{\cos^2(x)} - 5 = 0$$

$$2\sin^2(x) - 2 - 5\cos^2(x) = 0$$

Now, replace $\sin^2(x)$ with $1 - \cos^2(x)$:

$$2(1 - \cos^2(x)) - 2 - 5\cos^2(x) = 0$$

$$2 - 2\cos^2(x) - 2 - 5\cos^2(x) = 0$$

$$-7\cos^2(x) = 0$$

Now, solve for $\cos^2(x)$:

$$\cos^2(x) = \frac{0}{-7}$$

$$\cos^2(x) = 0$$

Now, solve for $x$ by taking the square root:

$$\cos(x) = \pm \sqrt{0}$$

$$\cos(x) = 0$$

This implies that $x$ is an angle where $\cos(x) = 0$. These angles are $x = \frac{\pi}{2} + n\pi$ and $x = \frac{3\pi}{2} + n\pi$, where $n$ is an integer.