I have no idea how to rearrange $$3\tan^2 (x) - \sec (x) - 1 = 0$$ to solve for $x$.
So far all I can do is rearrange it to form $$\frac{3\sin^2 (x) - \cos (x) - \cos^2 (x)}{\cos^2 (x)}$$
I don't know where to go from here or any other way to solve $x$
$\endgroup$ 13 Answers
$\begingroup$You have a good start. From there you have
$$\frac{3\sin^2 (x) - \cos (x) - \cos^2 (x)}{\cos^2 (x)} = 0\tag{1}$$
which is equivalent to $$3\sin^2 (x) - \cos (x) - \cos^2 (x) = 0\tag{2}$$
since any $x$ such that $\cos x = 0$ is not a solution of $(2)$. Note that $\sin^2x + \cos^2x = 1$, so $(2)$ becomes
$$3-\cos x-4\cos^2 x = 0.$$
Can you finish now?
$\endgroup$ $\begingroup$Hint:$\tan ^{2}x=\sec^{2}x -1$ so you only have to solve a quadratic in the variable $y=\sec x$.
Answer: $x=sec^{-1}(-1)$ or $x =sec^{-1}(\frac 4 3)$.
$\endgroup$ $\begingroup$We know that $\tan^2x=\sec^2x-1$ so our equation becomes \begin{align}3\tan^2(x)−\sec(x)−1&=0\\ 3(\sec^2(x)-1)-\sec(x) -1&=0\\ 3\sec^2(x)-3-\sec(x)-1&=0\\ 3\sec^2(x)-\sec(x)-4&=0\end{align}
Now we have a quadratic equation, where $y=\sec(x)$:
\begin{align}3y^2-y-4&=0\\ (3y-4)(y+1)&=0\end{align}
So now we know that either \begin{align}3\sec(x)-4&=0\\ x&=\sec^{-1}\left(\frac43\right)\end{align}
or \begin{align}\sec(x)+1&=0\\ x&=\sec^{-1}(-1)\\ &=\pi\end{align}
$\endgroup$
