Wikipedia:Reference desk/Archives/Mathematics/2010 May 15

= May 15 =

2cosx+2cos(2x)=0
Consider the function $$f(x)=2\sin x+\sin2x$$

Locate and classify the function's stationary points.

$$\frac{df(x)}{dx}=2\cos x+2\cos2x$$

$$0=2\cos x+2\cos2x$$

$$0=\cos x+\cos2x$$

$$0=\cos x+\cos^2x-\sin^2x$$

$$0=\cos x(1+\cos x-\sin x\cdot\tan x)$$

$$\cos x=0$$

$$x=\frac{\pi}{2}$$

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 $$0=\cos x+\cos^2x-\sin^2x$$ into a quadratic in terms of $$\cos x$$. --COVIZAPIBETEFOKY (talk) 11:55, 15 May 2010 (UTC)
 * $$0=\cos x+\cos^2x-(1-\cos^2x)$$
 * $$0=2\cos^2x+\cos x-1$$
 * $$\cos x = \frac{-1\pm\sqrt{1^2-4\cdot 2\cdot (-1)}}{2\cdot 2}=\frac{1}{2}, -1$$
 * $$x=\frac{\pi}{3}, \pi$$ —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 $$2\cos2x$$, 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)