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.

• 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 $2$.

• 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

• 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

To solve $(\sin x+\cos x)^{1+\sin 2x}=2$ when $−\pi\leq x\leq \pi$. Let $z=\sin x+\cos x$, then $z^2=1+\sin 2x$. Hence the equation is equivalent to $z^{z^2}=2$ which has the only possible solution $z=\sqrt{2}$.

Hence $\sin x+\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})=\sqrt{2}$.

Thus $\sin(x+\frac{\pi}{4})=1$, which implies $x+\frac{\pi}{4}=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.

Or using $z^2=1+\sin 2x$ with $z=\sqrt{2}$ gives $1+\sin 2x=2$

Hence $\sin 2x=1$ implies $2x=\frac{\pi}{2}$. Hence $x=\frac{\pi}4$.