1
$\begingroup$

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?

$\endgroup$

2 Answers 2

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 $$2\sqrt 2\cos(2x+45^\circ)=\sqrt 6\Rightarrow\cos(2x+45^\circ)=\frac{\sqrt 6}{2\sqrt 2}=\frac{\sqrt 3}{2}.$$

$\endgroup$
1
  • $\begingroup$ Thank you - was a silly mistake :) $\endgroup$
    – hohner
    Commented Oct 10, 2014 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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .