Need help with last bit of my equation: f(x)=8*sin(x)+4cos(x)*sin(x) 0<x<2*pi

[danishkid]

Hi, so I'm doing a project and I've figured most of it out
Very short:
I have to prove that theres an maximum point for this function:
f(x)=8*sin(x)+4cos(x)*sin(x) 0<x<2*pi

I've derived it and set it equal to zero but now I've come down to the last part. I have to solve this:
2cos(x)=-cos(2x)

Bear in mind that I can't use any math solving software, it has to be done by hand. I know what the answer is (1.19606) because I've made a graph of each side of the equation but how would you solve it by hand?

I would be so happy for any reply since this is extremely important for me to get right so thank you in advance

 

[Deleted member 4993]

Hi, so I'm doing a project and I've figured most of it out
Very short:
I have to prove that theres an maximum point for this function:
f(x)=8*sin(x)+4cos(x)*sin(x) 0<x<2*pi

I've derived it and set it equal to zero but now I've come down to the last part. I have to solve this:
2cos(x)=-cos(2x)

Bear in mind that I can't use any math solving software, it has to be done by hand. I know what the answer is (1.19606) because I've made a graph of each side of the equation but how would you solve it by hand?

I would be so happy for any reply since this is extremely important for me to get right so thank you in advance

2cos(x)=-cos(2x)

2cos²(x) + 2cos(x) - 1 = 0

Above is a quadratic equation - solve it using your favorite method. According to your problem statement, you do NOT need to solve for 'x', just need to prove that 'x' is real.

 

[danishkid]

2cos(x)=-cos(2x)

2cos²(x) + 2cos(x) - 1 = 0

Above is a quadratic equation - solve it using your favorite method. According to your problem statement, you do NOT need to solve for 'x', just need to prove that 'x' is real.

Wow, I didn't realize that. Makes me feel kind og stupid
Thank you so much for the help! If I may ask you a last question; how can I determine whether the point(s) are a minimum or maximum? Any easy method?

Edit: I do have to find a specifik value for x, but maybe I've done something wrong along the way. But thank you anyway

 

[HallsofIvy]

Wow, I didn't realize that. Makes me feel kind og stupid
Thank you so much for the help! If I may ask you a last question; how can I determine whether the point(s) are a minimum or maximum? Any easy method?

1. The first derivative test: If the derivative at x= a is 0, the derivative for x< a is negative, and the derivative for x> a is positive then x= a is a minimum. If the derivative for x< a is positive, and the derivative for x> a is negative then x= a is a minimum.
(In the first case, the graph is going down to the minimum, then up. In the second, the graph is going up to the maximum, then down.)

2. The second derivative test: If the derivative at x= a is 0 and the second derivative at x= a is positive, then x= a is a minimum. If the second derivative at x= a is negative. then x= a is a maximum.
(in the first case, the first derivative is going from negative to positive and in the second, from positive to negative.)

The second method requires that you find the second derivative but only have to evaluate it at one point. The first method doesn't require you to differentiate again but you need to be able to determine the sign of the first derivative over an interval.

 

[danishkid]

1. The first derivative test: If the derivative at x= a is 0, the derivative for x< a is negative, and the derivative for x> a is positive then x= a is a minimum. If the derivative for x< a is positive, and the derivative for x> a is negative then x= a is a minimum.
(In the first case, the graph is going down to the minimum, then up. In the second, the graph is going up to the maximum, then down.)

2. The second derivative test: If the derivative at x= a is 0 and the second derivative at x= a is positive, then x= a is a minimum. If the second derivative at x= a is negative. then x= a is a maximum.
(in the first case, the first derivative is going from negative to positive and in the second, from positive to negative.)

The second method requires that you find the second derivative but only have to evaluate it at one point. The first method doesn't require you to differentiate again but you need to be able to determine the sign of the first derivative over an interval.

Thank you so much, I didn't know any of these things so that was really helpful! I appreciate it a lot, both of you