0
$\begingroup$

How I solve the following equation for $0 \le x \le 360$:

$$ 2\cos2x-4\sin x\cos x=\sqrt{6} $$

I tried different methods. The first was to get things in the form of $R\cos(x \mp \alpha)$:

$$ 2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\ 2\cos2x-2\sin2x=\sqrt{6}\\ R = \sqrt{4} = 2 \\ \alpha = \arctan \frac{2}{2} = 45\\ \therefore \cos(2x + 45) = \frac{\sqrt6}{2} $$

which is impossible. I then tried to use t-substitution, where:

$$ t = \tan\frac{x}{2}, \sin x=\frac{2t}{1+t^2}, \cos x =\frac{1-t^2}{1+t^2} $$

but the algebra got unreasonably complicated. What am I missing?

$\endgroup$
1
$\begingroup$

Note that $$2\cos(2x)-2\sin (2x)=\sqrt{2^2+2^2}\cos(2x+\alpha)$$ where $$\cos\alpha=\sin\alpha=\frac{2}{2\sqrt 2}=\frac{1}{\sqrt 2}.$$

Hence, you'll have $$2\sqrt 2\cos(2x+45^\circ)=\sqrt 6\Rightarrow\cos(2x+45^\circ)=\frac{\sqrt 6}{2\sqrt 2}=\frac{\sqrt 3}{2}.$$

$\endgroup$
  • $\begingroup$ Thank you - was a silly mistake :) $\endgroup$ – hohner Oct 10 '14 at 15:36
1
$\begingroup$

You have made a mistake in the third step. $$ 2\cos2x-2(2\sin x\cos x)=\sqrt{6}\\ 2\cos2x-2\sin2x=\sqrt{6}\\ R = \sqrt{8} = 2\sqrt{2} \\ \alpha = \arctan -\frac{2}{2} = -\frac{\pi}{4}\\ \therefore \cos(2x - \frac{\pi}{4}) = \frac{\sqrt3}{2} $$

Hope this help.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.