Area Under Curve: Find Intersection Points & Area

[bonfire09]

Homework Statement

Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

Homework Equations

The sketch of the curve is given too which I uploaded.

The Attempt at a Solution

Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
0=2sin(6x)-1
sin(6x)=1/2
6x=sin^-1(1/2)
x=(sin^-1(1/2))/6

Then I am not sure what to do from here.

 

[HallsofIvy]

Homework Statement

Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

Homework Equations

The sketch of the curve is given too which I uploaded.

The Attempt at a Solution

Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)

Good up to here.

0=2sin(6x)-1

either 2sin(6x)- 1= 0 or cos(x)= 0

sin(6x)=1/2

Have you never learned that sin(\pi/6)= 1/2?

Draw an equilateral triangle with side length s. All three angles have measure \pi/3. Draw a perpendicular from one vertex to the opposite side forming a right triangle. That perpendicular bisects the angle and the opposite side. The angle at that vertex is (\pi/3)/2= \pi/6. The hypotenuse is a side of the original triangle and has length s. The "opposite side" has length s/2 so sin(\pi/6)= (s/2)/s= 1/2.

6x=sin^-1(1/2)
x=(sin^-1(1/2))/6

Then I am not sure what to do from here.

6x= \pi/6 so x= \pi/36

Of course, cos(x)= 0 for x= \pi/2.

Since sine and cosine are periodic, there are many solutions to these equations. Your graph should show which one you need to use.

 

[haruspex]

Sketch the regions enclosed by the given curves.
y = 3 cos 6x, y = 3 sin 12x, x = 0, x = π/12
Find the area as well.

Area or areas?

The sketch of the curve is given too which I uploaded.

Where?

Trouble finding intersection points
3cos(6x)= 3sin(12x)
cos(6x)=sin(2*6x)
cos(6x)=2sin(6x)cos(6x)
0=cos(6x)(2sin(6x)-1)
0=2sin(6x)-1

There's another solution, but I'll assume you already covered this in your sketch.

sin(6x)=1/2

What angle has sine equal to 0.5?
Assuming the idea is to find the total area enclosed, treating each individual area as positive, you need to break the range of integration at each intersection. (If you don't do that some areas will come out negative and cancel against others.)
Within each subrange, figure out which function is the greater. If in the range [a,b] f(x)>g(x) then you integrate f-g over that range.