Trig Identity Solutions: Solving csc^2(x/2) = 2secx | x Solutions

[Stanc]

Homework Statement

Need some help finding all solutions for x...

csc^2((x)/(2)) = 2secx

The Attempt at a Solution

Not sure what kind of approach to take but:

1/ sin^2(x/2) = 2/ cos x

From here Not sure what to do i tried cross multiplying and got cos x = 2sin^2(x/2) but got no idea from here... please help

 

[Stanc]

Does anyone have an explanation for this solution?? someone just sent it:

csc^2(x/2)=2secx
2/(1-cosx)=2secx
2=2secx(1-cosx)
1=secx-1
2=secx
π/3+2πk=x
5π/3+2πk=x

However i do think he's missing a squared? Can anyone explain this formula?

 

[rock.freak667]

Is your problem

cosec²(x/2) = secx

or

½cosec²(x)=secx


If it is the first one, then your double angle formulas for cos2x will help you out here greatly.

 

[Mark44]

Does anyone have an explanation for this solution?? someone just sent it:

csc^2(x/2)=2secx
2/(1-cosx)=2secx

The line above is incorrect on the left side. It should be
2/(1 - cos²(x/2)) = 2sec(x)

2=2secx(1-cosx)
1=secx-1
2=secx
π/3+2πk=x
5π/3+2πk=x

However i do think he's missing a squared? Can anyone explain this formula?

To solve this equation, convert everything into terms involving cosine. You need the double angle formula to convert cos(x) into cos(2 * x/2).

BTW, your title is misleading. You do not "solve" an identity - you prove that an equation is identically true (true for all or most values of the variable). What this seems to be is an equation to solve for specific values of x.

 

[SammyS]

Homework Statement

Need some help finding all solutions for x...

csc^2((x)/(2)) = 2secx

The Attempt at a Solution

Not sure what kind of approach to take but:

1/ sin^2(x/2) = 2/ cos x

From here Not sure what to do i tried cross multiplying and got cos x = 2sin^2(x/2) but got no idea from here... please help

I would be inclined to go from where you are here

cos(x) = 2sin^2(x/2)​

and use the double angle formula to convert cos(x) to a form having sin(x/2). In other words, think of cos(x) as cos(2(x/2)) .

Then you will have an expression involving only sin(x/2) .

 

[Stanc]

I would be inclined to go from where you are here

cos(x) = 2sin^2(x/2)​

and use the double angle formula to convert cos(x) to a form having sin(x/2). In other words, think of cos(x) as cos(2(x/2)) .

Then you will have an expression involving only sin(x/2) .

So basically change cosx into 1-sin^2(x/2) ??

 

[HallsofIvy]

So basically change cosx into 1-sin^2(x/2) ??

NO, cos x is NOT equal to 1- sin^2(x/2). It is equal to 1- sin^2(x). You could then use the identity sin(x)= 2sin(x/2)cos(x/2).

 

[Stanc]

The line above is incorrect on the left side. It should be
2/(1 - cos²(x/2)) = 2sec(x) To solve this equation, convert everything into terms involving cosine. You need the double angle formula to convert cos(x) into cos(2 * x/2).

BTW, your title is misleading. You do not "solve" an identity - you prove that an equation is identically true (true for all or most values of the variable). What this seems to be is an equation to solve for specific values of x.

Sorry for the title too
Sorry i do not follow why you have the 2 over in 2/(1-cos^2(x/2))

 

[Stanc]

NO, cos x is NOT equal to 1- sin^2(x/2). It is equal to 1- sin^2(x). You could then use the identity sin(x)= 2sin(x/2)cos(x/2).

Thanks, i wasnt too sure about that part... But isn't it cos^2x that is suppose to equal 1 - sin^2x?

 

[ehild]

Homework Statement

Need some help finding all solutions for x...

csc^2((x)/(2)) = 2secx

The Attempt at a Solution

Not sure what kind of approach to take but:

1/ sin^2(x/2) = 2/ cos x

From here Not sure what to do i tried cross multiplying and got cos x = 2sin^2(x/2) but got no idea from here... please help

Apply the half-angle formula: sin²(x/2)=(1-cosx)/2

ehild

 

[Stanc]

Apply the half-angle formula: sin²(x/2)=(1-cosx)/2

ehild

Yes, i can solve it with the half angle formula but was wondering if i could solve it without that formula...

 

[ehild]

You need to use either the half-angle formula or the double-angle one.

ehild

 

[Stanc]

You need to use either the half-angle formula or the double-angle one.

ehild

How would i approach it with double angle? I followed some of the steps and came with this

1/1-cos^2(x/2) = 2/cos2(x/2)

2-2cos^2(x/2) = cos2(x/2)

Is this right?

 

[SammyS]

So basically change cosx into 1-sin^2(x/2) ??

Not quite:

cos(x) = 1 - 2sin²(x/2)

 

[Stanc]

Not quite:

cos(x) = 1 - 2sin²(x/2)

Oh ya, sorry about that, so from here i have:

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

If i move the right side over i would get:

1-4sin^2(x/2) = 0

sin^2(x/2) = 1/4

Square root it:

sin(x/2) = 1/2?

I think that's correct, right?

 

[ehild]

sin(x/2) = ±1/2.

ehild

 

[Mark44]

Sorry for the title too
Sorry i do not follow why you have the 2 over in 2/(1-cos^2(x/2))

That 2 shouldn't be there. The line I was correcting had 2/(1 - cos(x)), and I brought that 2 along, not noticing that it was wrong as well.

 

[Stanc]

sin(x/2) = ±1/2.

ehild

Thanks., but the answers i get are pi/6 and 5pi/6 while the answers are pi/3 and 5pi/3

Anything i have to do?

 

[SammyS]

Thanks., but the answers i get are pi/6 and 5pi/6 while the answers are pi/3 and 5pi/3

Anything i have to do?

If x/2 = π/6 , then x = π/3 ...