Question

determine the max and min values of f(x)=2sinx+sin(2x) on the interval [0,2pi]

Answer 1

step 1: take the derivative

Answer 2

I don't know if @mariaad is here.
But I guessed at the end. I wonder if there is a simple way to solve that explicitly? Or is it double angle and all that...

Answer 3

i am here sorry!

Answer 4

for the derivative i got y= 2cosx + 2cos(2x)

Answer 5

i just don't know when i set it equal to 0 what to do

Answer 6

yeah it's easy with \[\cos(2x)=2\cos^2x-1\]so using that it will be quadratic in cosx

Answer 7

i dont understand

Answer 8

You got
y = 2cos(x) + 2cos(2x)

You can make the substitution
cos(2x) = 2cos^2 (x) - 1

So you have
y = 2cos(x) + 2(2cos^2 (x) - 1)
y = 4 cos^2 x + 2 cos(x) - 2

Which you can solve like a quadratic equation.