Square-Trigonometry Equations!

$\begin{aligned}\dfrac1{\sin^2{\theta}}-\dfrac1{\cos^2{\theta}}-\dfrac1{\tan^2{\theta}}-\dfrac1{\cot^2{\theta}}-\dfrac1{\sec^2{\theta}}-\dfrac1{\csc^2{\theta}}=&\ -3\\\csc^2{\theta}-\sec^2{\theta}-\cot^2{\theta}-\tan^2{\theta}-\cos^2{\theta}-\sin^2{\theta}=&\ -3\\(\csc^2{\theta}-\cot^2{\theta})-\sec^2{\theta}-\tan^2{\theta}-(\cos^2{\theta}+\sin^2{\theta})=&\ -3\\1-(1+\tan^2{\theta})-\tan^2{\theta}-1=&\ -3\\-1-\tan^2{\theta}-\tan^2{\theta}=&\ -3\\-2\tan^2{\theta}=&\ -2\\\tan^2{\theta}=&\ 1\\\tan{\theta}=&\ \pm1\end{aligned}$

Thus,

$\begin{cases}\tan{\theta}=1\Rightarrow\theta=\dfrac{\pi}4,\dfrac{5\pi}4\\\tan{\theta}=-1\Rightarrow\theta=\dfrac{3\pi}4,\dfrac{7\pi}4\end{cases}$

Hence, This equation has $\boxed4$ solutions in the interval $(0,2\pi)$ .

$\dfrac1{\sin^2{\theta}}-\dfrac1{\cos^2{\theta}}-\dfrac1{\tan^2{\theta}}-\dfrac1{\cot^2{\theta}}-\dfrac1{\sec^2{\theta}}-\dfrac1{\csc^2{\theta}}=-3$ Writing this in terms of $\sin{\theta}$ and $\cos{\theta}$ , we obtain $\dfrac1{\sin^2{\theta}}-\dfrac1{\cos^2{\theta}}-\dfrac{\cos^2{\theta}}{\sin^2{\theta}}-\dfrac{\sin^2{\theta}}{\cos^2{\theta}}-\cos^2{\theta}-\sin^2{\theta}=-3$ Subsitute all $\sin^2{\theta}$ with $x$ and $\cos^2{\theta}$ with $1-x$ , we obtain $\dfrac1{x}-\dfrac1{1-x}-\dfrac{1-x}{x}-\dfrac{x}{1-x}-(1-x)-x=-3$ $\dfrac{1-x}{x-x^2}-\dfrac{x}{x-x^2}-\dfrac{1-2x+x^2}{x-x^2}-\dfrac{x^2}{x-x^2}-1=-3$ $\dfrac{1-x-x-1+2x-x^2-x^2}{x-x^2}=-2$ $\dfrac{-2x^2}{x-x^2}=-2$ Multiplying on both sides, we get $-2x^2=-2x+2x^2$ $4x^2-2x=0$ Factoring yields $2x(2x-1)=0$ $x=0,\frac{1}{2}$ Recall that $\sin^2{\theta}=x$ , so $sin^2{\theta}=0, \frac{1}{2}$ , and $\sin{\theta}=0, \pm\dfrac{1}{\sqrt{2}}$ . However, $\sin{\theta}$ cannot equal $0$ because then $\sin^2{\theta}$ would be $0$ and $\dfrac{1}{\sin^2{\theta}}$ would be undefined. Therefore $\sin{\theta}$ must equal $\pm\dfrac1{\sqrt{2}}$ or $\dfrac{\sqrt{2}}{2}$ . Therefore $\theta$ must have $\boxed{4}$ solutions: $\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, and \dfrac{7\pi}{4}$ .