Question

sin 3x +cos 2x =0 find the general solution

Answer 1

cos(2x) = -sin(3x) 
= -cos(π/2 - 3x) 
= cos(π/2 + 3x) 
:.Either 
2n π + 2x = π/2 + 3x 
x = (2n - 1/2)pi 
or 
2n π - 2x = π/2 + 3x 
x = (2n - 1/2) π/5

Answer 2

Answer:

The solution are

$x = (2n - \frac{1}{2})\pi$

and $x = (2n - \frac{1}{2}) \frac{\pi}{5}$

Step-by-step explanation:

We have to find the general solution for the equation

$\sin 3x +\cos 2x =0$

⇒ $\cos(2x) = -\sin(3x)$

As cos Ф=sin(90-Ф)

We can write

-sin Ф=cos(90+Ф), as cos Ф is -ve in second quadrant.

$\cos 2x=\cos(\frac{\pi}{2} + 3x)$

The solution is

$2n\pi \pm 2x=\frac[\pi}{2}+3x$

Either  

$2n\pi + 2x=\frac[\pi}{2}+3x$

$x = (2n - \frac{1}{2})\pi$

or  

$2n\pi - 2x=\frac[\pi}{2}+3x$

$x = (2n - \frac{1}{2}) \frac{\pi}{5}$