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How do I solve this trig equation?

$$\cos 2x - \sin2x = \sqrt{3} \cos 4x$$

I have tried in different ways. But I can't get to a final answer.Please help.

My work is,

$$\cos 2x - \sin 2x = √3(\cos 2x - \sin 2x)(\cos 2x + \sin 2x)$$ $$⇔(\cos 2x - \sin 2x)((1-√3(\cos 2x + \sin 2x))= 0$$ Then, $$\tan 2x = 1 \ \ \ \text{or} \ \ \ (\cos 2x + \sin 2x)= \dfrac1{\sqrt3}$$

I don't know whether this way is correct. Please someone help for a better solution.

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    $\begingroup$ Show us the work from your different attempts. That will help us better understand any misunderstandings you might have, and what you may have overlooked. $\endgroup$
    – amWhy
    May 26, 2017 at 15:19
  • $\begingroup$ I have tried writing the cos 4x as (cos 2x)^2-(sin 2x)^2 $\endgroup$
    – Nimantha
    May 26, 2017 at 15:24
  • $\begingroup$ It should be $\cos^2 2x - \sin^2 2x$ $\endgroup$ May 26, 2017 at 15:25
  • $\begingroup$ actually, what you wrote is common short-hand and means precisely what the asker used. Nimantha: note that $(\cos 2x)^2 = \cos^2(2x)$. So you are indeed correct. $\endgroup$
    – amWhy
    May 26, 2017 at 15:27

2 Answers 2

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We need to solve that $$\cos2x-\sin2x=\sqrt3(\cos2x-\sin2x)(\cos2x+\sin2x),$$ which gives $\cos2x-\sin2x=0$, which is $x=\frac{\pi}{8}+\frac{\pi}{2}k,$ $k\in\mathbb Z$ or $$\cos2x+\sin2x=\frac{1}{\sqrt3},$$ which is $$\cos\left(2x-\frac{\pi}{4}\right)=\frac{1}{\sqrt6},$$ which gives $x=\frac{\pi}{8}\pm\frac{1}{2}\arccos\frac{1}{\sqrt6}+\pi k$.

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HINT:

Use $\cos2y=\cos^2y-\sin^2y$ in $\cos4x=\cos2(2x)$

Also, for $\sqrt3(\cos2x+\sin2x)=1,$ $$\cos\left(2x-\dfrac\pi4\right)=\dfrac1{\sqrt6}$$

$$\implies2x-\dfrac\pi4=2m\pi\pm\arccos\dfrac1{\sqrt6}$$ where $m$ is any integer.

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