Trig equations: solving csc(5x) = sin(7x) over the interval [0, 2pi]

[dreager]

I have never had a class make me feel so stupid. This is getting to me.

I am having a hard time solving trig equations over a given interval.

For example csc(5x)=sin(7x) ,[02pi].

I now that first I must make the equation = to zero.

so this would give me
csc(5x)-sin(7x)=0

Then I must use identities to change the equation.
sin(theta)=1/csc(theta).

so now I have
csc(5x)-1/csc(7x).

Multiply both sides by csc(7x) to get
csc(35x)=0??? This is where I am getting tripped up.

Or do square both sides n factor?

Any help would be great.
Thank you

 

[stapel]

I am having a hard time solving trig equations over a given interval.

For example, solving csc(5x) = sin(7x) over the interval [0, 2pi].

A good first step is to convert everything to sines and cosines. Since the cosecant is the reciprocal of the sine, you get:

. . . . .\(\displaystyle \dfrac{1}{\sin(5x)}\, =\, \sin(7x)\)

Of course, this means that x cannot be anything which will make sin(5x) equal to zero, because this would create a division-by-zero problem. So we'll need to remember to exclude all solutions to "5x = n*pi", where "n" is an integer.

I now that first I must make the equation = to zero.

so this would give me: csc(5x)-sin(7x)=0

Then I must use identities to change the equation: sin(theta)=1/csc(theta).

so now I have: csc(5x)-1/csc(7x).

In general, sines are better than cosecants.

Multiply both sides by csc(7x) to get: csc(35x)=0

How are you getting this? What identity did you apply? (You know, from back in algebra, that "f(2x)*f(3x)" is almost never equal to f(6x) or f(6x^2). So you must have done something else. But what?)

By the way, Wolfram Alpha gives a very complicated answer (here). Are you sure you copied the exercise correctly? Thank you!

 

[Deleted member 4993]

I have never had a class make me feel so stupid. This is getting to me.

I am having a hard time solving trig equations over a given interval.

For example csc(5x)=sin(7x) ,[02pi].

I now that first I must make the equation = to zero.

so this would give me
csc(5x)-sin(7x)=0

Then I must use identities to change the equation.
sin(theta)=1/csc(theta).

so now I have
csc(5x)-1/csc(7x).

Multiply both sides by csc(7x) to get
csc(35x)=0??? This is where I am getting tripped up.

Or do square both sides n factor?

Any help would be great.
Thank you

This problem has no-solution within the given domain.

 

[dreager]

SO after looing closer at this problem. I am now at the point where I have converted everything to Sin

1/sin(5x)=sin(7x). see that this as no solution because there is no x value within one rotation of the circle that will give me the solution. Am Imaking sense o am I missing something?

 

[stapel]

A good first step is to convert everything to sines and cosines. Since the cosecant is the reciprocal of the sine, you get:

. . . . .\(\displaystyle \dfrac{1}{\sin(5x)}\, =\, \sin(7x)\)

This problem has no-solution within the given domain.

SO after looing closer at this problem. I am now at the point where I have converted everything to Sin: 1/sin(5x)=sin(7x). see that this as no solution...

Okay, so you've correctly copied what was provided to you. Now, please show us what you have done to prove that there is no solution. Thank you!