+1911 Let $x$ be an acute angle such that $\tan x + \sec x = 2 \cot x.$ Find $\cos x$.
+1557 we get [(\tan x + \sec x)^2 = 4 \cot^2 x.]Then tan2x+2tanxsecx+sec2x=4cot2x. We can use the Pythagorean identitytan2x+1=sec2x to rewrite the left-hand side of this equation as [(\tan^2 x + 1) + 2 \tan x \sec x = \sec^2 x + 2 \tan x \sec x = 4 \cot^2 x.]Hence, tan2x+1=4cot2x,which simplifies tocot2x−tan2x=1.
Dividing both sides bycos2x,we getcsc2x−sec2x=sec2x,socsc2x=2sec2x. Then cscx=2secx,sosinx1=2cosx1. Squaring both sides and rearranging, we get \begin{align*} \sin^2 x &= 2 \cos^2 x\ \cos^2 x &= \frac{\sin^2 x}{2}\ \cos x &= \boxed{\pm \frac{\sqrt{2}}{2}}. \end{align*}Since xis acute,cosx must be positive, so cosx=sqrt(2)/2.
