Asked by Anonymous
Solve the equation algebraically for x wohere 0 </ x< 2 pi.
sin2 x = cos x
sin2 x = cos x
Answers
Answered by
bobpursley
sin^2 x=cosx ?
1-cos^2 x=cosx
0=cos^2 x + cosx -1
cosx=(-1+-sqrt(1+4))/2=-1/2 +-1/2 sqrt 5
x= arccos(-.5-.5sqrt5) or arccos(-.5+.5sqrt5) one of these does not work
then x=51.8deg
check my math
1-cos^2 x=cosx
0=cos^2 x + cosx -1
cosx=(-1+-sqrt(1+4))/2=-1/2 +-1/2 sqrt 5
x= arccos(-.5-.5sqrt5) or arccos(-.5+.5sqrt5) one of these does not work
then x=51.8deg
check my math