# Solution of the trigonometric equation $\sin5x =\cos2x$

The question is to find the general solution of $\sin5x = \cos2x$ I did the following:

$\sin5x-\cos2x=0$

$\implies\sin5x-\sin(\frac{\pi}2-2x)=0$

$\implies\cos(\frac{3x}2 +\frac{\pi}4)\sin(\frac{7x}2 - \frac{\pi}4)=0$

And I got the answers as:

$(4n + 1)\frac{\pi}{14} = x$ (Which is correct)

And

$(4n+1)\frac{\pi}6 =x$ (Which apparently is wrong and the correct answer is given to be $x=-(4n-1)\frac{\pi}6$ Could somebody please tell me where I went wrong?

• Whats the difference between the answer the you got and the correct one? They look the same to me. – José Carlos Santos Jan 15 '18 at 14:52
• In my answer it is $(4n + 1)$ and in the given answer it's $(1 - 4n)$ – Le Connoisseur Jan 15 '18 at 14:53
• With $n\in\mathbb Z$, right?! And the other answer is $-4n+1$, with $n\in\mathbb Z$. What's the difference? – José Carlos Santos Jan 15 '18 at 14:54
• Oh, yeah. I didn't think of that. facepalm At least now I'll think of it the next time. Thanks! – Le Connoisseur Jan 15 '18 at 14:56

I think it's better after your first step to use the following. $$5x=\frac{\pi}{2}-2x+2\pi k,$$ where $k\in\mathbb Z$ or

$$5x=\frac{\pi}{2}+2x+2\pi k.$$

Also you can write your equation as

$${\rm Im}[e^{i5x}]={\rm Re}[e^{\pm i2x}]$$

Hence, using the geometry of the complex plane and renaming $v:=e^{i5x}$ and $w:=e^{\pm i2x}$, you need to find all $x$ such that $v\cdot(- i)=w$, that is

$$e^{i5x}e^{- i(\pi/2+2k\pi)}=e^{i(5x-\pi/2+2k\pi)}=e^{\pm i2x}\implies\begin{cases} 3x=\frac\pi2+2k\pi\\7x=\frac\pi2+2k\pi\end{cases}\implies x=\begin{cases}\pi\left(\frac16+\frac23k\right)\\\pi\left(\frac1{14}+\frac27k\right)\end{cases}$$

for all $k\in\Bbb Z$.