# Solving $2\cot 2x\cos2x = 1-\sin 2x$

How would I solve the following trigonometric equation?

$$2\cot 2x\cos2x = 1-\sin 2x$$

I got to this stage: $2\cos^22x = \sin2x - \sin^22x$

How do I continue?

## 5 Answers

$$2cot2xcos2x=1-sin2x\\ \frac { \cos ^{ 2 }{ 2x } }{ \sin { 2x } } =\frac { 1-\sin { 2x } }{ 2 } \\ 2\left( 1-\sin ^{ 2 }{ 2x } \right) +\sin ^{ 2 }{ 2x } -\sin { 2x } =0\\ \sin ^{ 2 }{ 2x } +\sin { 2x } -2=0\\ \sin { 2x } =\frac { -1\pm 3 }{ 2 } \\ \sin { 2x\neq -2 } ,\sin { 2x } =1\\ \sin { 2x } =1\Rightarrow 2x=\frac { \pi }{ 2 } +2n\pi \Rightarrow \\$$ so the final answer is:

$$x=\frac { \pi }{ 4 } +n\pi$$

That leads to: $2(1-t^2) = t^3-t^4, t = \sin(2x)$. Can you see common factor here?

after applying the addition formulas we get $$2\cos(x)^4-2\cos(x)^2-\sin(x)\cos(x)+1=0$$ we converting this in $\tan(x/2)$ we get$$\left( {t}^{4}-2\,{t}^{3}+2\,{t}^{2}+2\,t+1 \right) \left( {t}^{2}+2 \,t-1 \right) ^{2} =0$$ with $$t=\tan(x/2)$$

Let $u=\cos 2x$. The equation becomes: $$2\frac{u^2}{\sqrt{1-u^2}}=1-\sqrt{1-u^2}$$ or $$2u^2=\sqrt{1-u^2}-1+u^2$$ which leads to $$(u^2+1)^2=1-u^2$$ that is $$u^4+3u^2=0$$ which has only one solution: $u=0$, that is, $x\equiv \pi/4\pmod{\pi/2}$

• You are assuming that $sin(2x)\geq0$ when you write $\sin(2x)=\sqrt{1-u^2}$ – D.L. Sep 7 '15 at 19:39

How about this:

$$\frac{2\cos 2x}{\sin 2x}\cos 2x = 1 -\sin 2x$$

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

Use the fact that $\cos^2 2x = 1 -\sin^2 2x$ to express the equation in terms of $\sin 2x$

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

$$(\sin 2x +2)(\sin 2x - 1) = 0$$

I'm sure you can finish it from here.

• Yes, I did this part but I don't know how to continue – LiziPizi Sep 7 '15 at 19:19
• @Chinny84, thanks, typo. – John_dydx Sep 7 '15 at 19:21
• error between the second and the third line (plus/minus error, I think) – D.L. Sep 7 '15 at 19:32
• @D.L., sry, I don't really see any error here. – John_dydx Sep 7 '15 at 19:34
• Sorry, I hadn't refreshed.... Now, it's correct (I had the expression with the $3.\sin^2(2x)$ on my screen. – D.L. Sep 7 '15 at 19:37