<s>[INST] Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$ 
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$

Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}$
Why is this solution wrong? Is there more than one $x$ value for $\sin x - 1=0$ or $(\sqrt{2} \cos x +1)=0$ in the given interval? How do I know that?
 [/INST] 
Can you find out where did you miss? Look at the blue graph. It's the graph of $y=\sqrt{2} \cos x + 1$. So, the root of the blue graph should be $x=\pi \pm \dfrac{\pi}{4} \Rightarrow x = \dfrac{3\pi}{4}, \dfrac{5\pi}{4}$.
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