Confused on applications of solving $\cos \frac x2 - \sin x = 0$ Why can't $\cos\frac{x}{2} - \sin x = 0 $ be transformed into being $\cos(x) - \sin(2x) = 0$? It works in application to achieve the half angle formulas for $\sin x$, $\cos x$, and $\tan x$, why doesn't it work here?
 A: Hint:
Let $x \in \mathbb{R}$; $\cos (x/2) - \sin x = 0$ iff $\cos (x/2)[ 1 - 2\sin (x/2) ] = 0$; $\cos (x/2)[ 1 - 2\sin (x/2) ] = 0$ iff $\cos (x/2) = 0$ or $\sin (x/2) = 1/2$. 
A: Your (not-reasonable) confusion stems from the difference in how identities and equations use their variables.


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*In an identity such as $\cos 2 x = 2\cos^2 x - 1$, the "$x$" stands for "any number". (That's the special nature of identities!) Thus, replacing "$x$" by, say, "$x/2$" wherever it appears "works" ---that is, yields a still-true statement--- because $x/2$ is also "any number".

*In an equation such as $\cos(x/2) - \sin x=0$, the "$x$" stands for "a particular number" ... or, more precisely, "a member of a particular set of numbers". Replacing "$x$" by "$2x$" wherever it appears changes the meaning of the symbol "$x$" to "a member of some other set of numbers". For instance, the original equation has $x=\pi$ as a solution. In the "$2x$" version, $\cos x - \sin 2x = 0$, the corresponding solution is $x=\pi/2$. The "$x$" is being used differently.
Now, in the second situation, if you keep careful track of things, solving the $2x$-equation and recalling that "these new $x$s are half(!) the size of the original $x$s, so we need to convert back, ...", then everything's fine. However, such an argument can be very hard to follow by your audience (or even yourself, if you get distracted in the middle of work). The better, safer strategy is to introduce an auxiliary variable, as in: "Let's define $y:=x/2$, so that we can write the equation as $\cos y - \sin 2y = 0$." Then, when you find that $y=\pi/2$ is a solution of the transformed equation, there's no confusion over which $x$ solves which equation; moreover, the conversion to a solution of the original equation is crystal clear to everyone: $x=2y=2(\pi/2) = \pi$.
A: $\cos x=2\cos^2\dfrac x2-1=2\sin^2x-1=-\cos2x$
So, we need $\sin2x=-\cos2x\iff\tan2x=-1$
$2x=n\pi-\dfrac\pi4\  \ \ \  (1)$ where $n$ is any integer
$\cos\dfrac x2=\sin x=\cos\left(\dfrac\pi2-x\right)$
$\implies\dfrac x2=2m\pi\pm\left(\dfrac\pi2-x\right)$ where $m$ is any integer
Check for both the signs if  they conform to $(1)$
