# Solve $\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$ [closed]

Solve $$\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$$ Now I did not understand how can i solve that.

I have tried substituting $\cos(2x)=\cos^2(x)−\sin^2(x)$ and$\,$ $\sin(2x)=2\sin(x)\cos(x)$,

the equation is now $(1−\sqrt2\sin(x))(\cos^2(x)−\sin^2(x)+2\sin(x)\cos(x))=\frac12$

Help Required

Thanks

## closed as off-topic by Henrik, Rohan Shinde, Namaste, Saad, Xander HendersonMar 26 '18 at 0:34

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• @TrầnThúcMinhTrí Not anyone is an expert in English. The OP truly hasn't posed an attempt on a solution but judging his English knowledge isn't correct. – Rebellos Mar 25 '18 at 12:00
• No, I mean I just don't like how somebody ask a question that says something too short like "Help needed"; "Help required", etc. There is a page that mentions about writing good questions. – user061703 Mar 25 '18 at 12:02
• I have tried $\cos 2x= \cos ^{2}x- \sin ^{2}x$ and $\sin 2x= 2\sin x\cos x$, the equation is $\left ( 1- \sqrt{2}\sin x \right )\left ( \cos^{2}x- \sin ^{2}x+ 2\sin x\cos x \right )= \frac{1}{2}$ – user530708 Mar 25 '18 at 12:04
• Can you edit my queston? – user530708 Mar 25 '18 at 12:05
• Perhaps the $t=tan(x)$ substitution would work? – Sorfosh Mar 25 '18 at 16:17

## 2 Answers

Multiplying by $1+\sqrt 2 \sin x$ on both sides and using $1-2 \sin^2 x = \cos 2x$ we get

$\cos 2x (\cos 2x + \sin 2x ) = \dfrac{1+\sqrt 2 \sin x}{2}$

or $1+\cos 4x + \sin 4x = 1+\sqrt 2 \sin x$

or $\dfrac{\sin 4 x + \cos 4x}{\sqrt 2} = \sin x$

or $\sin \left(4x+\dfrac{\pi}{4} \right) = \sin x$ which can be easily solved

• From there, $\arcsin$ both sides, right? – ericw31415 Mar 25 '18 at 20:53
• I tried to use Wolfram alpha but your answer were not true. The equation has 5 roots like John Glenn. Thanks for your idea very much! – user530708 Mar 26 '18 at 0:27

You can substitute: $$y=2x\Rightarrow x=\frac y2$$ $$\left ( 1- \sqrt{2}\sin \frac y2 \right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$$ Remember that: $$\sin\frac\theta2=\pm\sqrt{\frac{1-\cos\theta}2}$$ So you get: $\left ( 1- \sqrt{2} (\pm\sqrt{\frac{1-\cos\theta}2})\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$, which simplifies to: $$\left ( 1 \pm\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$$ Solving for the first case:$\left ( 1 -\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$ \begin{align} \cos y-(\cos y) \sqrt{1-\cos (y)}+\sin y-(\sin y) \sqrt{1-\cos (y)}&=\frac{1}{2}\\ -(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}&=\frac{1}{2}-\cos y-\sin y\\ \left(-(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}\right)^2&=\left(\frac{1}{2}-\cos y-\sin y\right)^2\\ \end{align}

We then evaluate each side: \begin{align} \left(-(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}\right)^2&=\cos ^2 y-\cos ^3 y+2 (\cos y) (\sin y)-2 \left(\cos ^2 y\right) (\sin y)+\sin ^2 y-(\cos y) \left(\sin ^2 y\right)\\ \left(\frac{1}{2}-\cos y-\sin y\right)^2&=y \sin ^2-y \sin +y \cos ^2-y \cos +2 (y \sin ) (y \cos )+\frac{1}{4}\\ \end{align}

Continuing the computation, we get: \begin{align} -\frac{1}{4}+\cos y-\cos ^3 y+\sin y-2 \left(\cos ^2 y\right) (\sin y)-(\cos y) \left(\sin ^2 y\right)&=0\\ -1+4 (\cos y)-4 \left(\cos ^3 y\right)+4 (\sin y)-8 \left(\cos ^2 y\right) (\sin y)-4 (\cos y) \left(\sin ^2 y\right)&=0\\ -1-4 (\sin y)+8 \left(\sin ^3 y\right)&=0\\ (2 (\sin y)+1) \left(-1-2 (\sin y)+4 \left(\sin ^2 y\right)\right)&=0\\ \end{align} We get the correct answers as: $$y=\begin{cases} 2 \pi n+\frac{7 \pi }{6}\qquad{n \in \mathbb{Z}}\\ 2 \pi n+\frac{3 \pi }{10}\\ 2 \pi n+\frac{11 \pi }{10}\\ 2 \pi n+\frac{19 \pi }{10}\\ \end{cases}$$

And since $x=\frac y2$, then the first set of solutions for $x$ will be: $$x=\begin{cases} 2 \pi n+\frac{7 \pi }{12}\qquad{n \in \mathbb{Z}}\\ 2 \pi n+\frac{3 \pi }{20}\\ 2 \pi n+\frac{11 \pi }{20}\\ 2 \pi n+\frac{19 \pi }{20}\\ \end{cases}$$

The second case:$\left ( 1 +\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$ returns the following answers: $$y=\begin{cases} 2 \pi n-\frac{ \pi }{6}\qquad{n \in \mathbb{Z}}\\ 2 \pi n+\frac{7 \pi }{10}\\ \end{cases}$$ However,the second answer for $x$ does not hold for the original equation. And thus, our final solution set contains: $$\therefore x=\begin{cases} 2 \pi n-\frac{ \pi }{12}\qquad{n \in \mathbb{Z}}\\ 2 \pi n+\frac{7 \pi }{12}\\ 2 \pi n+\frac{3 \pi }{20}\\ 2 \pi n+\frac{11 \pi }{20}\\ 2 \pi n+\frac{19 \pi }{20}\\ \end{cases}$$

• John Glenn, thanks! – user530708 Mar 26 '18 at 0:23