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I want to find the general solution of the following equation for $x \in \mathbb{R}$:

$$\cos (5x) = \sin (x)$$

I know it might sound silly, but I don't know how to bring $5x$.

Did
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Rat Rod
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  • See if http://math.stackexchange.com/questions/643217/how-can-i-solve-sinx-sin2x inspires you. – mlc Apr 01 '17 at 17:09

5 Answers5

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HINT: $$\cos(5x)=\sin(\frac{\pi}{2}-5x)\\\sin(\frac{\pi}{2}-5x)=\sin x$$

kingW3
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$cos5x=cos(\pi/2-x)$ so this implies $5x=\pi/2-x+k2\pi$, or $5x=-(\pi/2-x)+k2\pi$. Now you can solve for $x$ (and thus $5x$)

imranfat
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    Slightly better than kingW3's almost simultaneous answer with the same approach, as the eveness of cosine makes it easier to express the two angles in each period with matching cosines. – Paul Sinclair Apr 01 '17 at 21:32
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using the Addition formulas we obtain $$\sin(5x)=16\,\sin \left( x \right) \left( \cos \left( x \right) \right) ^{4}- 12\,\sin \left( x \right) \left( \cos \left( x \right) \right) ^{2}+ \sin \left( x \right) $$

Dr. Sonnhard Graubner
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Hint:

$$\cos 5x = \sin x$$

$$\sin 90^°- 5x = \sin x$$

$\sin a= \sin b$ then $a=n\pi + (-1)^nb $ where $n∈N$

Fawad
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$$ \cos 5x = \cos ^5(x)-10 \sin ^2(x) \cos ^3(x)+5 \sin ^4(x) \cos (x) $$

trigs

One solution $$x = \pm \frac{\pi}{4}$$

dantopa
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