the question is:

Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.

Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$

similarly solving the above equation it comes out to be:

$$\sqrt2 u^2-u-\sqrt2=0=(u-\sqrt2)(\sqrt2u+1)=0 \implies \sin x+\cos x=\sqrt2 \quad(1)$$ and $$\sin x+\cos x= \frac{-1} {\sqrt2}\quad (2)$$

so solving the results differently i got the answers:

$$x=2n\pi + \frac{\pi}4, 2n\pi +\frac{7\pi}{12}, 2n\pi-\frac{\pi}{12}$$

but the answers are:

$$x=2n\pi + \frac{\pi}4, 2n\pi -\frac{5\pi}{12}, 2n\pi+\frac{11\pi}{12}$$

I divided the eq(2) by $\sqrt2$

  • $\begingroup$ Aren't the answers same. You just have to put the next 'n' to change from say $-\frac{5\pi}{12}$ and $\frac{7\pi}{12}$ $\endgroup$ – Shailesh Aug 10 '15 at 8:03
  • $\begingroup$ Almost, but there is $2\pi$, not $\pi$ next to $n$. $\endgroup$ – orion Aug 10 '15 at 8:11
  • $\begingroup$ It seems to me that both your answer and the answer from the book(?) are wrong. Harish result below is correct. $\endgroup$ – mickep Aug 10 '15 at 8:15
  • $\begingroup$ $\sin x-\cos x$ doesn't match $u$. $\endgroup$ – Yves Daoust Aug 10 '15 at 8:45

Notice, we have $$\sin x-\cos x-2\sqrt2\sin x\cos x=0$$ $$\sin x-\cos x=2\sqrt2\sin x\cos x$$ $$\sin x-\cos x=\sqrt2\sin 2x$$ $$\frac{1}{\sqrt2}\sin x-\frac{1}{\sqrt2}\cos x=\sin 2x$$ $$\sin 2x=\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}$$ $$\sin 2x=\sin \left(x-\frac{\pi}{4}\right)$$ Writing the general solution , we get $$2x=2n\pi+\left(x-\frac{\pi}{4}\right)\iff x=\frac{(8n-1)\pi}{4}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{(8n-1)\pi}{4}}}$$

or $$2x=2n\pi+\pi-\left(x-\frac{\pi}{4}\right)\iff x=\frac{(8n+5)\pi}{12}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{(8n+5)\pi}{12}}}$$

Where, $n$ is any integer.


This could be helpful: \begin{align} \sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x&=-\sqrt2\Big(\sin(x-\frac{\pi}{4})+\sin(2x)\Big)\\ &=-\frac{\sqrt2}{2}\cos(\frac{\pi}{8}-\frac{3}{2}x)\sin(\frac{\pi}{8}+\frac{1}{2}x) \end{align}


The equation can be rewritten



$$x-\frac\pi4=2x+2k\pi\text{ or }x-\frac\pi4=\pi-2x+2k\pi,$$

$$x=\frac{24k-3}{12}\pi\text{ or }x=\frac{8k+5}{12}\pi.$$ The second formula covers the first, so that $$\color{green}{x=\frac{8k+5}{12}\pi}.$$


my hints: $$\sin \theta=\sin \alpha\iff \theta=2k\pi+\alpha\ \text{or}\ \theta =(2k+1)\pi-\alpha$$


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