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Find the number of solutions to the equation $\sin(2x) = \cos(3x)$ If $0\le x \le 2\pi$.

My approach: $$\cos(\frac{\pi}{2}-2x) = \cos3x$$ $$3x = 2n\pi \pm (\frac{\pi}{2}-2x)$$

taking positive, I got $$x=\frac{2n\pi}{5}+\frac{\pi}{10}$$

Negative may not be acceptable due to the range of the x.

Plugging in the values of n, I am only getting 4 solutions. But the answer given in my book is 6. Can anyone help me?

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2 Answers 2

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Solving: $$\sin(2x) = \cos(3x)$$ $$\cos(\frac{\pi}{2}-2x) = \cos3x$$ $$3x = 2n\pi \pm (\frac{\pi}{2}-2x)$$ We get: $$(1)\space 3x=2n\pi+{\pi\over2}-2x$$ $$(2)\space 3x=2n\pi-{\pi\over2}+2x$$ We make it look more clean: $$(1)\space x={2n\pi\over5}+{\pi\over10}$$ $$(2)\space x=2n\pi-{\pi\over2}$$ From (1) you find that $n=0,1,2,3,4$. From (2) you find that $n=1$
Edit: If you plug the values you get:
$x_1={\pi\over10},$
$x_2={\pi\over2},$
$x_3={9\pi\over10},$
$x_4={13\pi\over10},$
$x_5={17\pi\over10},$
$x_6={3\pi\over2}$
which are valid solutions(and unique for $x\in[0,2\pi]$)

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For '+', $$x=\dfrac{(4n+1)\pi}{10}$$

We need $$0\le\dfrac{(4n+1)\pi}{10}\le2\pi\iff0\le4n+1\le20\implies n=0,1,2,3,4$$

For negative we have one, right?

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  • $\begingroup$ @MeesdeVries, Thanks $\endgroup$ Nov 17, 2017 at 9:16

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