# Solutions for $\sin\left(3x-\frac{\pi}{2}\right)=\cos\left(x-\frac{\pi}{3}\right)$

How to solve the latter, anyone knows some hints or info on solving the trigonometric equation?

$$\sin\left(3x-\frac{\pi}{2}\right)=\cos\left(x-\frac{\pi}{3}\right)$$

I've tried the identities for addition and subtraction, e.g.

$$\sin(x-y)=\sin x\cos y - \sin y \cos x$$ and $$\cos(x-y)=\cos x \cos y+\sin x \sin y$$ but it didn't get me anywhere.

• use the identity $\cos x = \sin (\frac {\pi}{2} - x)$ – Doug M Jun 20 '17 at 22:41
• Which identities did you use for addition and subtraction, and show us how you used them, to help us spot an error where you may have gotten stuck. – Namaste Jun 20 '17 at 23:03
• regular ones : sin(x-y)=sinxcosy - sinycosx and cos(x-y)=cosxcosy+sinxsiny – Sal Khanov Jun 20 '17 at 23:09

Use the identity $\sin(\theta-\pi/2) = -\cos\theta$,
\begin{align*} \sin\left(3x - \frac\pi2\right) &= \cos\left(x-\frac\pi3\right)\\ -\cos3x &= \cos\left(x-\frac{2\pi}6\right)\\ 0&= \cos 3x + \cos\left(x-\frac{2\pi}6\right)\\ 0&= 2\cos\left(2x-\frac{2\pi}{12}\right)\cos\left(x+\frac{2\pi}{12}\right)\\ 2x-\frac{2\pi}{12} &= \frac{2\pi}2 n + \frac{2\pi}{4} &&\text{or}&x+\frac{2\pi}{12} &= \frac{2\pi}2 n + \frac{2\pi}4\\ x &= \frac{2\pi}{4}n + \frac{2\pi}6&&\text{or}&x &= \frac{2\pi}2 n + \frac{2\pi}{6} \end{align*}
Since $$\sin\left(3x-\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}-\left(3x-\frac{\pi}{2}\right)\right)=\cos\left(\pi-3x\right),$$ we need to solve $$\cos\left(\pi-3x\right)=\cos\left(x-\frac{\pi}{3}\right),$$ which gives $$\pi-3x=x-\frac{\pi}{3}+2\pi k,k\in\mathbb Z,$$ which is $$x=\frac{\pi}{3}+\frac{1}{2}\pi k, k\in\mathbb Z$$ or $$\pi-3x=-\left(x-\frac{\pi}{3}\right)+2\pi k,k\in\mathbb Z,$$ which is $$x=\frac{\pi}{3}+\pi k, k\in\mathbb Z$$ and we got the answer: $$\left\{\frac{\pi}{3}+\frac{1}{2}\pi k| k\in\mathbb Z\right\}.$$