How to solve $\sin(2x)\cdot(\sin(x)+\cos(x))=\sqrt2$

Question

I need help in solving equation: $\sin(2x)\cdot(\sin(x)+\cos(x))=\sqrt2$ I tried to

Answer 1

Hint:

We get $$\sqrt2=\sqrt2\sin(x+\pi/4)\sin2x$$

$$\implies\sin2x=\sin(x+\pi/4)=\pm1$$(why?)

Answer 2

Write your equation in the form $$\frac{1}{2} \left(-\sqrt{2} \sin \left(\frac{\pi }{4}-3 x\right)+\sqrt{2} \sin \left(x+\frac{\pi }{4}\right)-2 \sqrt{2}\right)=0$$ after this use $$\sin(x)-\sin(y)=2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \cos \left(\frac{x}{2}+\frac{y}{2}\right)$$

Answer 3

Taking square both sides, we have

$$\sin^22x(\cos^2x+\sin^2x+2\sin x \cos x )=\sin^22x(1+2\sin x \cos x )$$

$$\sin^2 2x(1+\sin2x)=2$$

Let $\sin2x=t$, where $-1\le t \le 1$.

$$t^3+t=2$$

$$(t-1)(t^2+t+2)=0$$

Since $t^2+t+2>0$ for all $t\in [-1,1]$, $t$ must be 1.

$$\sin2x=1$$

implies $\displaystyle2x=n\pi +(-1)^n\frac{\pi}{2}$ where $n\in \mathbb{Z}.$

Therefore, $$x=\frac{n\pi}{2} +(-1)^n\frac{\pi}{4} $$

where $n\in \mathbb{Z}.$

Answer 4

With $x=t+\dfrac\pi 4$,

$$-\cos(2t)\frac1{\sqrt2}(\sin t+\cos t+\cos t-\sin t)=\sqrt 2$$

or

$$\cos(2t)\cos t=-1.$$

The only possibility is $\cos t=-1$, $t=(2k+1)\pi$, i.e. $$x=(2k+1)\pi-\dfrac\pi4.$$

Answer 5

Avoid squaring as it immediately introduces When do we get extraneous roots?

Set $\sqrt2y=\sin x+\cos x,2y^2=?$

So, we have $$\sqrt2y(2y^2-1)=\sqrt2$$

By observation or using Rational root theorem,

one of real roots of the last equation is $1$

The rest two roots are imaginary