# Homework Help: Finding two vlaues for trig function

1. Apr 11, 2012

### rohan03

1. The problem statement, all variables and given/known data

I need to find at least two values of x in the interval [0,∏/2]
for which f(x)= sin(x)+sin(2x)+sin(3x)=1

3. The attempt at a solution

now this is what my understanding is and this is what I have done

sin(x)+sin(2x)+sin(3x)=1
sin(x)+sin(2x)+sin(3x)-1

Now by calculation f(0) = sin(0)+sin(2x0)+sin(3x0)-1 =-1

and f(∏/2) = sin(∏/2)+sin(2x∏/2)+sin(3x∏/2)-1
that then given me = 1+0+-1 -1 =-1

but I know this is not right- as I don't get negative to positive solution to show that there is atleast two values of x for which f(x) is true!!!
I think I have lost the plot somewhere!!!!!

can someone look at this and tell me whats wrong and guide me to correct procedure

2. Apr 11, 2012

### SammyS

Staff Emeritus
If the following is true,
then
Which says that f(x) - 1 = 0 .

So the rest is wrong.
Use the angle addition identity for sin(x+2x) which is sin(3x) .

Similarly, for sin(x), you can think of x as being 2x - x .

3. Apr 12, 2012

### rohan03

ok I am still lost

4. Apr 12, 2012

### rohan03

Thank you I found similar example which cleared all my misunderstanding

Last edited: Apr 12, 2012
5. Apr 12, 2012

### rohan03

thia gives me
2sin x + 2 sin x cos x - 2sin³ x + 2 cos² x sin x.

6. Apr 12, 2012

### SammyS

Staff Emeritus
f(x) - 1 = sin(x)+sin(2x)+sin(3x) -1
= sin(2x-x)+sin(2x)+sin(2x+x) -1

= sin(2x)cos(x)-sin(x)cos(2x) + sin(2x) + sin(2x)cos(x)+sin(x)cos(2x) -1

= ...​

7. Jun 15, 2012

### Fizzman

To expand further on this do you take say sin(2x) and break it down or do you start substituting values for x?

8. Jun 15, 2012

### SammyS

Staff Emeritus
Hello Fizzman. Welcome to PF !

Two of the terms cancel.

Collect terms that have sin(2x), then see what happens if you expand sin(2x) .

9. Jun 15, 2012

### Fizzman

Thanks for the welcome.

Ok so doing that I get

2sin(2x)cos(x) +sin(x) -1

Do I substitute for x or is further expansion needed?

If I expand the sin(2x)

2sin(2x)cos(x)+2sin(x)cos(x)-1

Last edited: Jun 15, 2012
10. Jun 15, 2012

### SammyS

Staff Emeritus
After looking at this problem for a while, it doesn't appear that it works out very nicely.

I get f(x)=sin(2x)cos(x)-sin(x)cos(2x) + sin(2x) + sin(2x)cos(x)+sin(x)cos(2x) -1

= sin(2x)(2cos(x) + 1) - 1

You can expand sin(2x) out & get a term with cos2(x) --- which can be changed to 1 - sin2(x) --- but there still remains one term with cos(x) in it.

That messes up the whole works!