Solve cos t = -tan(x) times tan(x) for x, and state when a solution exists.

Step-by-step solution

1. 1

   Collapse the repeated factor. The right-hand side is a product of a quantity with itself:

   $\tan(x)\tan(x)=\tan^2(x)\quad\Rightarrow\quad \cos t=-\tan^2(x)$

2. 2

   Use the range of $\tan^2$ to find when a solution can exist at all. A square of a real number is never negative, so $\tan^2(x)\ge 0$ and therefore $-\tan^2(x)\le 0$. The left-hand side must match, giving the solvability condition

   $\cos t\le 0\quad\Longleftrightarrow\quad t\in\left[\tfrac{\pi}{2}+2k\pi,\ \tfrac{3\pi}{2}+2k\pi\right]$

   If $\cos t>0$ there is no real $x$ whatsoever — checking this first saves solving an impossible equation.

3. 3

   Isolate the square.

   $\tan^2(x)=-\cos t$

   Under the condition above the right side is $\ge 0$, so taking a square root is legitimate.

4. 4

   Take both square roots.

   $\tan(x)=\pm\sqrt{-\cos t}$

   Keeping both signs is essential: dropping the negative branch loses half the solution family.

5. 5

   Invert the tangent over its full period. The tangent has period $\pi$ (not $2\pi$) and $\arctan$ is an odd function, so the two branches merge into one formula:

   $x=\pm\arctan\left(\sqrt{-\cos t}\right)+k\pi,\qquad k\in\mathbb{Z}$

6. 6

   Check two representative values of $t$. If $\cos t=0$ (say $t=\tfrac{\pi}{2}$) the formula gives $x=k\pi$, and indeed $-\tan^2(0)=0=\cos\tfrac{\pi}{2}$ ✓. If $t=\pi$ then $\cos t=-1$, so $\tan^2 x=1$ and $x=\pm\tfrac{\pi}{4}+k\pi$; substituting, $-\tan^2\!\left(\tfrac{\pi}{4}\right)=-1=\cos\pi$ ✓.