Let $ a=sin(x+\frac{\pi}{4})$. If $ x\in [0,\frac{\pi}{2}]$, then $ \frac{\sqrt 2}{2}\le a\le 1$ and $ 2-sin2x=3-2a^2$. Therefore $ 2a^3-3a+1=0$. One root is $ a=1$ or $ x=\frac{\pi}{4}+2k\pi$. Other roots are $ 2a^2+2a-1=0\to a=\frac{-1\pm \sqrt 3}{2}$. But $ \frac{\sqrt 3 -1}{2}<\frac{\sqrt 2}{2}$.

I simplified the given expression to get $ sin^3x + cos^3x$ = $ 1/\sqrt 2$ And then let $ l = sinx + cos x$ and $ m = sinx cos x$ Edit: Here is my proof. We know $ l^2 - 2m = 1$ and $ l(1-m)$ = $ 1/\sqrt 2$ So we get $ l = 1/\sqrt 2(1-m)$ So then if write out and expand we get $ 4m^3 - 6m^2 + 1$ = $ 0$ And then solved to get $ 4m^3 - 6m^2 + 1$ = $ 0$ which has the roots $ 1/2$,$ 1 - \sqrt 3/2$,$ 1 + \sqrt 3/2$ But observe that the maximum of m which is $ sinx cos x$ is $ 1/2$ So m cannot equal $ 1 + \sqrt 3/2$ and since x is in the first quadrant both $ sinx$ and $ cosx$ are positive So m cannot be $ 1 - \sqrt 3/2$ which is negative. So m is $ 1/2$ We then get l is $ \sqrt 2$ Solving we get the only solutions are $ sinx = 1/ \sqrt 2$ which implies the only solution is $ x = \pi/4$ $ QED$ Edit: Dang Rust beat me to it.

Hey Rust, how did you get $ 2-\sin(2x)=3-2a^2$??

hello, applying the addition formulas we get $ \sqrt{2}\sin(x)+\sqrt{2}\cos(x)-\sqrt{2}\sin^2(x)\cos(x)-\sqrt{2}\sin(x)\cos^2(x)-1=0$. Converting in to $ \tan(\frac{x}{2})$ we get ${ 2\,{\frac {\tan \left( 1/2\,x \right) \sqrt {2}}{1+ \left( \tan \left( 1/2\,x \right) \right) ^{2}}}+{\frac { \left( 1- \left( \tan \left( 1/2\,x \right) \right) ^{2} \right) \sqrt {2}}{1+ \left( \tan \left( 1/2\,x \right) \right) ^{2}}}-4\,{\frac { \left( \tan \left(1/2\,x \right) \right) ^{2}\sqrt {2} \left( 1- \left( \tan \left( 1/2 \,x \right) \right) ^{2} \right) }{ \left( 1+ \left( \tan \left( 1/2\,x \right) \right) ^{2} \right) ^{3}}}-2\,{\frac { \left( 1- \left(\tan \left( 1/2\,x \right) \right) ^{2} \right) ^{2}\sqrt {2}\tan \left( 1/2\,x \right) }{ \left( 1+ \left( \tan \left( 1/2\,x \right) \right) ^{2} \right) ^{3}}}-1=0}$ Substituting $ \tan(\frac{x}{2})=t$ factoring and simplifying we have $ (t^4-2t^3+2t^3\sqrt{2}-2t-2t\sqrt{2}-1)(-t-1+\sqrt{2})^2=0$ Only $ t=\sqrt{2}-1$ fulfilles our equation and it gives $ x=\frac{\pi}{4}$. Sonnhard.

It's not difficult to let it be like ${{\sin }^{3}}x+{{\cos }^{3}}x=\frac{\sqrt{2}}{2}$ . Then by $Holder$ , we get $(1+1){{({{a}^{3}}+{{b}^{3}})}^{2}}\ge {{({{a}^{2}}+{{b}^{2}})}^{3}}$. let $a$ to be $\sin x$ and $b$ to be $\cos x$ ,we can notice that the $min$ of ${{\sin }^{3}}x+{{\cos }^{3}}x$ is just $\frac{\sqrt{2}}{2}$. so we get that $x=\frac{\pi }{4}$ is the only root when $x\in [0,\frac{\pi }{2}]$.

(2-2sinxcosx)(sinxcos45⁰+cosxsin45⁰)=1 √2(1-sinxcosx)(sinx+cosx)=1 (1-sinxcosx)(sinx+cosx)=1/√2 (sin²x+cos²x-sinxcosx)(sinx+cosx)=1/√2 sin³x+cos³x=1/√2 (sin³x+√(1-sin²x)³)²=1/2 1-3sin²xcos²x=1/2 sin⁶x+cos⁶x=1/2 sin³x+cos³x=1/√2 => sinxcosx=0 =>x=π/4