# Help in Solving a Trigonometric Equation

Solve the equation $$\left(\sin x + \cos x\right)^{1+\sin(2x)} = 2$$ when $-\pi \le x \le \pi$ .

I have tried to use $\sin (2x) = 2\sin x \cos x$ identity but I this doesn't lead me to a conclusion.

I will appreciate the help.

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Hint: $\sin\dfrac\pi4=\dfrac1{\sqrt2}$ and $\sqrt2^2=2$. –  Lucian Jun 22 '14 at 4:32
I appreciate that Lucian. I think, I got it now. Thanks! –  Kushashwa Ravi Shrimali Jun 22 '14 at 4:36

Hint: We have $\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)$. It is not easy for a small positive power of this to be $2$.

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So, $(\sqrt{2} \sin (x+ \pi/4) )^{1+ \sin 2x} = 2$ –  Kushashwa Ravi Shrimali Jun 22 '14 at 4:28
Yes, and since the sine always has absolute value $\le 1$, we need the exponent to be $2$, and $\dots$. –  André Nicolas Jun 22 '14 at 4:30
Check your expression for $x=\frac{\pi}{4}$ –  Claude Leibovici Jun 22 '14 at 4:33
You can say that the absolute value of $\sin x+\cos x$ is between $0$ and $\sqrt{2}$. So we need $1+\sin 2x=2$ and $|\sin x+\cos x|=1$. From that you can know everything. –  André Nicolas Jun 22 '14 at 4:36
You are welcome. –  André Nicolas Jun 22 '14 at 4:48

Setting $\displaystyle x+\frac\pi4=y, \sin2x=\sin2\left(y-\frac\pi4\right)=-\sin\left(\frac\pi2-2y\right)=-\cos2y$

$\displaystyle(\cos x+\sin x)^{1+\sin2x}=(\sqrt2\sin y)^{2\sin^2y}=\left((\sqrt2\sin y)^2\right)^{\sin^2y}=(2\sin^2y)^{\sin^2y}$

Now, $\displaystyle0\le\sin^2y\le1$

So, $\displaystyle(2\sin^2y)^{\sin^2y}$ will be $=2$ if $\displaystyle\sin^2y=1\iff\cos y=0\iff y=(2n+1)\frac\pi2$ where $n$ is any integer

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Well, I think for the last statement you made that $\cos^2 y = 0$ and $\cos^2 y = 1$ is a little bit doubtful. –  Kushashwa Ravi Shrimali Jun 22 '14 at 4:44
While, if y equals $\cfrac{\pi}{4}$ then, $\cos^2 y \ne 0$ and $\cos^2 y \ne 1$. –  Kushashwa Ravi Shrimali Jun 22 '14 at 4:45
@KushashwaRaviShrimali, Where is the doubt? We need $\sin^2y=1$ and $\cos^2y=1$ at the same time, right? –  lab bhattacharjee Jun 22 '14 at 4:46
Oh, yeah, I got it now.. Sorry! –  Kushashwa Ravi Shrimali Jun 22 '14 at 4:49
Another way to see that $\sin^2y=1$ is the only solution is to notice that: $(2\sin^2y)^{2\sin^2y}=z^z=2^2,z\ge 0$. –  mike Jun 22 '14 at 5:49