Find the absolute maximum and absolute minimum values of f on the given interval [0, pi/2]. f(t) = 2 cos t + sin 2t?

F ( t ) = 2 cos t + sin 2 t
f ` ( t ) = - 2 sin t + 2 cos 2 t
- 2 sin t + 2 cos 2 t = 0  / : 2
- sin t  + cos 2 t = 0
- sin t + cos² t - sin² t = 0
- sin t + 1 - sin² t - sin² t = 0
- 2 sin² t - sin t + 1 = 0    /· ( -1 )
2 sin² t + sin t - 1 = 0
u = sin t

sin t = 1/2,  t = π/6  ( another solution is t = 3π/2 ∉ [ 0, π/2 ] )
f ( π/6 ) = 2 cos π/6 + sin π/3 = 2 ·√3/2 +√3/2 = 3√3/2
f ( 0 ) = 2 cos 0 + sin 0 = 2 · 1 + 0 = 2
f ( π/2 ) = 2 cos π/2 + sin π/2 = 0 + 0 = 0 
f min : ( π/2,  0 )
f max: ( π/6, 3√3/2 )