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if I have the following trig equation:

$$\sin(10x) = \cos(2x)$$

I take the following steps to solve it:

  • I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{\pi}{2} - a\right) = \sin\left(\frac{\pi}{2} + a\right) = \cos(a)$;
  • Let's say that I have chosen $\sin\left(\frac{\pi}{2} + 2x\right)$, the equation becomes:

$$\sin(10x) = \sin\left(\frac{\pi}{2} + 2x\right).$$

  • Then, I know that:

$$\sin(f(x)) = \sin(g(x)) \Leftrightarrow f(x) = g(x) + 2\pi n, n \in \mathbb{Z} \lor f(x) = (\pi - g(x)) + 2\pi k, k \in \mathbb{Z}$$

This means that (for $f(x) = 10x$ and $g(x) = \frac{\pi}{2} + 2x$):

$$\sin(10x) = \cos(2x) \Leftrightarrow 10x = \frac{\pi}{2} + 2x + 2\pi n, n \in \mathbb{Z} \lor 10x = (\pi - (\frac{\pi}{2} + 2x)) + 2\pi k, k \in \mathbb{Z}$$

Solving, I get the following results:

$$x_{1} = \frac{\pi}{16} + \frac{\pi}{4}n,\,\,x_{2} = \frac{\pi}{24} + \frac{\pi}{6}k,\,\,\,\,n,k \in \mathbb{Z}$$

Now, are there any other methods for solving such equations or could this one be just fine?

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  • $\begingroup$ This is rhe most efficient method. $\endgroup$ – André Nicolas Jul 8 '15 at 21:55
  • $\begingroup$ draw extremely careful graphs of $\cos 2x$ and $\sin 10x$ for, say, $0 \leq x \leq 2 \pi,$ and see if the intersections agree with your calculations printablepaper.net/category/graph $\endgroup$ – Will Jagy Jul 8 '15 at 21:55
  • $\begingroup$ @Zach466920 Thanks! $\endgroup$ – user3019105 Jul 8 '15 at 21:57
  • $\begingroup$ @AndréNicolas All right, got it! $\endgroup$ – user3019105 Jul 8 '15 at 21:57
  • $\begingroup$ @WillJagy Are you saying that there is an error? $\endgroup$ – user3019105 Jul 8 '15 at 21:57
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You may prefer to transform the sine into cosine: $$ \cos\left(\frac{\pi}{2}-10x\right)=\cos(2x) $$ This splits into two: $$ \frac{\pi}{2}-10x=2x+2k\pi $$ or $$ \frac{\pi}{2}-10x=-2x+2k\pi $$ The trick is that $\cos\alpha=\cos\beta$ if and only if $\alpha=\beta+2k\pi$ or $\alpha=-\beta+2k\pi$ (with integer $k$).

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Otherwise as you ask could be as follows (although it is almost the same).

You have $$sin (f(x))-sin (\frac\pi2-g(x))=0$$ so from the identity

$$sin(a)-sin(b)=2sin\frac {a-b}{2}cos\frac{a+b}{2}$$

it follows $$2sin\frac{(f(x)+g(x)- \frac{\pi}{2})}{2}cos\frac{(f(x)-g(x)+\frac{\pi}{2})}{2}=0$$ hence $$f(x)+g(x)=2n\pi+\frac\pi2$$ and $$f(x)-g(x)=n\pi-\frac\pi2$$

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  • $\begingroup$ Another good example! $\endgroup$ – user3019105 Jul 9 '15 at 5:56

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