$$\cos^2(t)-\sin^2(t)=0$$
Using trig identity, we can write it equal to
$\cos(2t) = 0$ where I get $2t = \frac{\pi}{2}, \frac{3\pi}{2}$ which means $t = \frac{\pi}{4}, \ \frac{3\pi}{4}$
or (without trig identity)
$$\cos^2(t)=\sin^2(t)$$ where I get $t = \frac{\pi}{4},\ \frac{3\pi}{4}, \ \frac{5\pi}{4},\ \frac{7\pi}{4}$
What am I doing wrong?
$\cos(2t) = 0$ implies that $2t = \pi(n+1/2) $ where $n= 0, \pm 1, \pm 2, ... $ since $\cos(\pi(n+1/2)) = 0$ for those values of $n$. You're simply missing a few of these solutions above. In the same way your neglecting other solutions when you solve without trig identity. Does this answer your question?
Method$\#1:$ $$2t=(2m+1)\dfrac\pi2$$ where $m$ is any integer
Method$\#2:$
$$\cos^2t=\sin^2t\implies\sin^2t=1-\sin^2t$$
$$\iff\sin^2t=\dfrac12=\sin^2\dfrac\pi4$$
Now use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
If $\sin\left(t-\dfrac\pi4\right)=0, t-\dfrac\pi4=n\pi$ where $n$ is any integer
What if $\sin\left(t+\dfrac\pi4\right)=0?$
More generally, if $$\sin^2x=\sin^2A\iff\cos^2x=\cos^2A\iff\tan^2x=\tan^2A$$
Can you prove $x=r\pi\pm A$ where $r$ is any integer
Alternative solution:
$$\cos^2(t)-\sin^2(t)=0$$
$$\sin^2(t)=\cos^2(t)$$
$$\tan^2(t)=1$$
$$\tan(t)= \pm 1$$
thus $$t= \frac{π}4,\frac{3π}4,\frac{5π}4,\frac{7π}4$$
NOTE:
keep attention when from here $$\sin^2(t)=\cos^2(t)$$ dividing by $\cos^2(t)$ and obtain $$\tan^2(t)=1$$ you are assuming that $\cos^2(t) \neq0$ that's a correct assumption here because $t=\pm\frac{π}2$ do not satisfy the original equation.
