How do you solve $cos(x-pi/2)=sqrt3/2$ for $-pi<x<pi$ ?

Question

4949 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T09:19:15

$color(blue)(pi/3,(2pi)/3)$

Identity:

$color(red)bb(cos(A-B)=cosAcosB+sinAsinB)$

$cos(x-pi/2)=cos(x)cos(pi/2)+sin(x)sin(pi/2)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=cosx(0)+sinx(1)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=sinx$

$:.$

$sinx=sqrt(3)/2$

$x=arcsin(sinx)=arcsin(sqrt(3)/2)=>x=pi/3,(2pi)/3$

$color(blue)(pi/3,(2pi)/3)$

How do you solve cos(x-pi/2)=sqrt3/2cos(x-pi/2)=sqrt3/2 for -pi<x<pi-pi<x<pi?