0

Sec(x/2) = cos(x/2)

I worked on this and got here...

(Let (x/2) = u)

Cosu - secu = 0

Cosu(1 - Sec^2u) = 0

Cosu[ -1(-1 + sec^2u)] = 0

Cosu(-tan^2u) = 0

So, the solutions would be:

x = pi + 4pik, 3pi + 4pik, 0 + 2pik but the problem is that the first two (pi + 4pik and 3pi + 4 pik)end up making the original equation have an undefined term (sec(x/2)). Is this simply because I went out of terms of the original equation? If so does this mean that every time I go out of terms of the original I must check the answers? This is confusing me a lot because usually you don't have to check answers unless you square both sides. Please help me out! Thank you.

King Squirrel
  • 2,607

1 Answers1

0

Probably the first case you will meet in algebra where you have to check because you may have introduced extra solutions is, as you have said, when you square both sides. But there are many, many more which you will meet as you progress. Here is just one example - of course my final answer is wrong!

Problem. Solve $x=2x$.

Solution. We have $$\eqalign{x=2x\quad &\Rightarrow\quad \sin x=\sin2x\cr &\Rightarrow\quad \sin x=2\sin x\cos x\cr &\Rightarrow\quad \sin x=0\ \hbox{or}\ \cos x={\textstyle\frac{1}{2}}\ ,\cr}$$ so the solutions are $x=k\pi$ or $x=\pm\frac{\pi}{3}+2k\pi$.

It is worth while giving some thought to exactly what is meant by the statement

$x=a$ is a solution of the equation $LHS=RHS$.

It does not mean, as many people believe, that if the equation is true then $x=a$. It actually means the converse,

if $x=a$ then the equation is true.

So if you begin with the equation you cannot possibly (in a strict logical sense) find the solutions. You can only find some $x$ values which might be solutions, and then you have to check to see that they really are solutions.

Hope this helps!

David
  • 82,662