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May 15

2cosx+2cos(2x)=0

Consider the function $f(x)=2\sin x+\sin 2x$
Locate and classify the function's stationary points.

$\frac{df(x)}{dx}=2\cos x+2\cos 2x$
$0=2\cos x+2\cos 2x$
$0=\cos x+\cos 2x$
$0=\cos x+{\cos }^{2}x-{\sin }^{2}x$
$0=\cos x(1+\cos x-\sin x\cdot \tan x)$
$\cos x=0$
$x=\frac{\pi }{2}$
doesn't work because tan is undefined at pi/2 --Alphador (talk) 10:17, 15 May 2010 (UTC)

Pi/2 is a potentially false solution since in the case of x=pi/2 dividing by cos x is dividing by zero. So this solution has to be tested. Setting x=Pi/2 into 2cos x+2cos 2x=2cos Pi/2+2cos pi=-2Taemyr (talk) 11:17, 15 May 2010 (UTC)

Convert $0=\cos x+{\cos }^{2}x-{\sin }^{2}x$ into a quadratic in terms of $\cos x$. --COVIZAPIBETEFOKY (talk) 11:55, 15 May 2010 (UTC)

$0=\cos x+{\cos }^{2}x-(1-{\cos }^{2}x)$

$0=2{\cos }^{2}x+\cos x-1$

$\cos x=\frac{-1\pm \sqrt{1^2-4\cdot 2\cdot (-1)}}{2\cdot 2}=\frac{1}{2},-1$

$x=\frac{\pi }{3},\pi$ —Preceding unsigned comment added by 220.253.221.60 (talk) 12:14, 15 May 2010 (UTC)

The derivative of sin(2x) is not cos(2x). 76.229.218.70 (talk) 18:35, 15 May 2010 (UTC)

It is $2\cos 2x$, as the OP has correctly written. -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)

$-\pi /3$ is another solution, and all solution are of course $+2k\pi$. In fact, all solutions can be written succinctly as $(2k+1)\pi /3$. -- Meni Rosenfeld (talk) 18:42, 15 May 2010 (UTC)