CACULUS
posted by RAYMOND on .
THE VALUES OF X THAT ARE SOLUTIONS TO THE EQUATION [COSX]^2 = SIN2X IN THE INTERVAL 0 TO PI??

(cosx)^2 = 2sinxcox
(cosx)^2  2sinxcosx = 0
cosx(cosx  2sinx) = 0
cosx = 0 or cosx = 2sinx
if cosx=0 then x = pi/2
if cosx = 2sinx
sinx/cosx = 1/2
tanx = 1/2
x = .46365