Solving the Equation sin(x) + sqrt(3)cos(x) = 1

[Moose352]

For some reason, I seem to be unable to algebraically solve this equation:

sin(x) + sqrt(3)cos(x) = 1

Any help would be appreciated.

 

[Hurkyl]

You need to combnie the LHS into a single trig function.

 

[Moose352]

Never mind, LHS means left hand side.

Yes, I know I need to convert the left side into the same trig function. That is what I'm having trouble with.

 

[Hurkyl]

All righty.

Suppose the equation was of the form:

<br /> \cos \frac{\pi}{5} \sin x + \sin \frac{\pi}{5} \cos x = 1<br />

Would you be able to solve for x?

 

[Moose352]

Yes, but I don't know how exactly that is applied here.

 

[Hurkyl]

(I should've mentioned that there will be a couple steps to this)


Ok. pretend for a moment that you could solve the equations:

cos y = 1
sin y = &radic;3

Then would you be able to solve the equation:

sin x + &radic;3 cos x = 1

 

[matt grime]

There is a general formula for this, usuallr referred to as rsin(theta + x)

but here, have you thought about multiplying everything by the same number so you get something akin to Hurkyl's example (think of some obvious values of cos sin etc involving sqrt(3))?

 

[Moose352]

I'm sorry, but still nope :(

 

[Hurkyl]

So you know how to solve the equation:

cos y sin x + sin y cos x = z

for x, if you know what y and z are.


Now, if I want to solve the equation

A sin x + B cos x = z

and I know that

A = cos y
and
B = sin y

Then can you solve this equation for x?

 

[Moose352]

Hmm, I think I figured it out. Tell me if I am right:

cos(y) = z
sin(y) = z*sqrt(3)

So y = tan^-1(sqrt(3)) = pi/3

So

sin(x)cos(y) - cos(x)sin(y) = 1z
sin(x-y) = 1z
x-y = sin^-1(.5)

and then solve for x?

Thanks a lot

 

[Moose352]

is there any significance to the value z (in my previous post) always seeming to equal 1/sqrt(A^2 + B^2)?

 

[Hurkyl]

Well, what does \sin^2 x + \cos^2 x equal?

 

[Moose352]

That makes sense! I can't believe I didn't figure this problem out myself.

Thanks a lot for the help.