Draw the graph of `y=2cosx + sin2x`

Explanation

We have `y=f(x) = 2cosx+ sin2x`
Clearly, domain of the function is R.
Also `f(x)` is periodic with period `2pi`.
So we need to draw the graph for the interval for the interval `[0,2pi]`.
`f(0) =0`
`f(x)=0`
`therefore 2cosx+2sincosx=0`
`therefore 2cosx(1+sinx)=0`
`therefore cosx=0` or `sinx=-1`
`therefore x=pi//2` or `x=(3pi)//2`
`f^(')(x) = -2sinx+2cosx2x`
`=-2(2sin^(2)x+sinx-1)`
`=-2(2sinx-1)(sinx+1)`
`f^(')(x) =0`
`therefore sinx=1//2` or `sinx=-1`
`therefore x=pi//6, (5pi)/6` or `x=(3pi)/2`
`f(pi//6) = 2cos(pi//6) + sin(pi//3)`
`=(3sqrt(3))/2`
`f(5pi//6) = 2cos(5pi//6+sin(5pi//3))`
`=-3sqrt(3)/2`
`f(3pi//2)=0`
So important points on graph paper as shown in the following figure.

From the points A to B, f(x) increases, from points B to D, f(x) decreases intersecting the x-axis at C, from points D to F, f(x) increases intersecting the x-axis at E.
Further `f^('')(x)=-2 cosx-4sin2x`
`f^('')(x) =0` at `x=(3pi//2)`, hence it is the point of inflection.
Since functions have period `2pi`, the graph of `y=f(x)` for `x in R` is as shown in the following figure.