Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $2$.

share|cite|improve this answer
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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.