So this is the problem:
$$\sin2x=\sqrt2\cos x$$

$$\cos x(2\sin x-\sqrt2)=0$$ My question is, how can I come with this to the conclusion that, according to my answer sheet, $\cos x=0$ and $\sin x=\frac{\sqrt2}{2}$ Is it just plugging in something until it works?

  • 2
    $\begingroup$ It's just the general fact that $$ab=0\iff a=0\vee b=0$$ $\endgroup$ – user228113 Feb 12 '17 at 19:41
  • $\begingroup$ If you want an algebraic explanation, it's due to the fact that $\Bbb R$ is an integral domain. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 12 '17 at 19:42
  • $\begingroup$ Note that the answer should be $\cos x=0$ OR $\sin x=\frac{\sqrt2}{2},$ not "and." Also, when you write $\cos x(2\sin x-\sqrt2)=0$ someone might think you mean $\cos (x(2\sin x-\sqrt2))=0,$ although that's clearly not the correct equation. To make it clearer, you could write $(\cos x)(2\sin x-\sqrt2)=0$ or $(2\sin x-\sqrt2) \cos x=0.$ $\endgroup$ – David K Feb 12 '17 at 19:48

Recall in Algebra you solved problems like $(x-2)(x+3)=0$? You used the Zero Product Property, which @G.Sassatelli stated. So for my example you get $x-2=0$ or $x+3=0$, and solve.

For your equation, we have $(\cos x)\left(2\sin x-\sqrt{2}\right)=0$. We use the Zero Product Property to get $\cos x=0$ or $2\sin x-\sqrt{2}=0$. This second equation, upon solving for $\sin x$, results in $\sin x=\sqrt{2}/2$.


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