Question 794887
[0,2π]
solve this equation: {{{ cosx=sin2x }}}.
cosx=sin2x
cosx-sin2x=0
cosx-2sinxcosx=0
cosx(1-2sinx)=0
cosx=0
x=π/2, 3π/2
..
1-2sinx=0
2sinx=1
sinx=1/2
x=π/6, 5π/6
..
x=π/6, π/2, 5π/6, 3π/2