Easy way : go to desmos.com and plot.

Manual way :

since you are asked to label the coordinates of all the stationary points, find them first.

y = sin x (1 + cos x)
= sin x + sin x cos x
= sin x + ½sin 2x

dy/dx = cos x + cos 2x

when dy/dx = 0,

cos x + cos 2x = 0
cos x + 2cos²x - 1 = 0
(2cos x - 1)(cos x + 1) = 0
cos x = ½ or cos x = -1

x = π/3, 5π/3 or x = π

(We are only interested in 0 ≤ x ≤ 2π)

Sub these values (they are the special angles) and you will find that :

y = ¾√3 , -¾√3 or y = 0

So your turning points are :

(π/3, ¾√3), (π, 0) and (5π/3, -¾√3)

Next, find all the x-intercepts.

When y = 0,

sin x (1 + cos x) = 0

sin x = 0 or cos x = -1

x = 0 , π , 2π or x = π

(We are only the interested in 0 ≤ x ≤ 2π. The above are the special values if you recall the sin x and cos x functions


So the coordinates of the intercepts are
(0,0) , (π,0) and (2π,0)

Lastly, check the nature of the turning points (π/3, ¾√3), (π, 0) and (5π/3, -¾√3)

We know the coordinates of the x-intercepts :(0,0) , (π,0) and (2π,0)

Logically,

① Since y increases from 0 at x = 0 to ¾√3 at x = π/3, and then decreases back to 0 at x = π,

(π/3, ¾√3) is a maximum point.

② Since y decreases from 0 at x = π to -¾√3 at x = 5π/3, and then increases back to 0 at x = 2π,

(5π/3, -¾√3) is a minimum point.

③ putting the two pieces of info together, we know that (π,0) is a point of inflection.

You can now plot the curve and its modulus easily.

The original y = sin x (1 + cos x) is anti-symmetric about x = π.

The modulus y = |sin x (1 + cos x)| is symmetric about x = π since you have to flip the right side (which is negative) up.




ALTERNATIVELY,

You can either use the first derivative method (check values of dy/dx very close to either side of each turning point)

or second derivative method (check the sign of d²y/dx²)



If using the second derivative method :

Example :

d²y/dx² = -sinx - 2sin2x

Sub x = π/3, d²y/dx² = -√3 / 2 - √3 = -3√3 / 2
(Negative, so maximum point)

Thanks so much! Really appreciate the detailed explanations!

Welcome