Reviewing Trig I come across this problem :
$\text{Solve for all real $x$ such that } 2\sqrt{2} \cos\left(\frac{x}{2}\right)=\cos(x) + 2.$
The first thing I did was use the cosine half-angle identity get this look…
$$2\sqrt{2}\sqrt{\frac{\cos(x)+1}{2}}=\cos(x)+2$$
I then squared both sides of the equation (I believe this is an error but can't pinpoint why).
The rest looks like this :
$$8\frac{\cos(x)+1}{2}=\cos^2(x)+4\cos(x)+4$$
$$4\cos(x)+4=\cos^2(x)+4\cos(x)+4$$
$$0=\cos^2(x)$$
Then $x = \left\{\pm \frac{\pi}{2}+2\pi n \ \middle|\ n \in\mathbb Z \right\}$
Now, I believe this set does contain all $x$ which satisfy the orignial equation, but it definitely contains invalid solutions as well.
What specifically did I do wrong here and, if possible, are there any hard and fast rules about when algebraic operations on trig equations will change the solution set?
Best Answer
If you are trying to solve an equation of the type $f(x)=g(x)$, it is perfectly fine to do $f^2(x)=g^2(x)$. The solutions of the first equation will also be solutions of the second one. But there is a real possibility of creating new ones. An extreme case of this is the equation $x=-x$, whose only solution is $x=0$. But every real number is a solution of the equation $x^2=(-x)^2$.
In your specific case, take all solutions that you got and check which ones are solutions of the original equation.