How do you solve $4 sin (x + π) = 2 cos (x + \frac{π}{2}) + 2$ $4sin(x+pi)=2cos(x+pi/2)+2$ ?

Question

2406 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-14T09:27:02

$x = 2 n π - \frac{π}{2}$ $x=2npi-pi/2$

We can solve the equation $4 sin (x + π) = 2 cos (x + \frac{π}{2}) + 2$ $4sin(x+pi)=2cos(x+pi/2)+2$ ,

by using the identities $sin (x + π) = - sin x$ $sin(x+pi)=-sinx$ and $cos (x + \frac{π}{2}) = - sin x$ $cos(x+pi/2)=-sinx$

Hence $4 sin (x + π) = 2 cos (x + \frac{π}{2}) + 2$ $4sin(x+pi)=2cos(x+pi/2)+2$ can be written as

$- 4 sin x = - 2 sin x + 2$ $-4sinx=-2sinx+2$ or

$- 4 sin x + 2 sin x = 2$ $-4sinx+2sinx=2$ or

$- 2 sin x = 2$ $-2sinx=2$ or

$sin x = - 1 = s o n (- \frac{π}{2})$ $sinx=-1=son(-pi/2)$

Hence $x = 2 n π - \frac{π}{2}$ $x=2npi-pi/2$

How do you solve 4sin(x+pi)=2cos(x+pi/2)+24sin(x+π)=2cos(x+π2)+24sin(x+pi)=2cos(x+pi/2)+2?