0
$\begingroup$

I just ran into this question for an admission test...

How many solutions does the equation:

$$(1+\sec\theta)(1+\csc\theta)=0$$

have for $\theta \in [-\pi , \pi]$?

My trial so far:

$$2\cos^{2}\left(\frac{\theta}{2}\right)=1+\cos\theta$$

Divide both sides by $\cos\theta$ and I get:

$$1+\sec\theta=2\frac{\cos^{2}\left(\frac{\theta}{2}\right)}{\cos\theta}$$

Substituting and doing some algebra, I get that the solution would be $\sin\theta=-1$, which gives me $\theta=-\frac{\pi}{2}$.

However the correct solution is that there are no solutions in the interval...

What am I missing here? Any help is appreciated

Thanks in advance

$\endgroup$
2
$\begingroup$

The points where $\sec \theta = -1$ are exactly the points where $\csc \theta$ is undefined, and likewise for the other term. This is a consequence of the identity $\sin^2 x + \cos^2 x = 1$, so that $\cos x = -1 \implies \sin x = 0$.

$\endgroup$
  • $\begingroup$ Got it! So the correct procedure would be to check in the original equation if the possible root(s) won't cause any problems? $\endgroup$ – bertozzijr Sep 5 '17 at 19:14
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$ – user296602 Sep 5 '17 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.