Re: solving trig equation

On Sep 16, 12:43 am, José Carlos Santos <jcsan...@xxxxxxxx> wrote:
On 16-09-2007 6:57, conrad wrote:





I have the following problem:

2*cos(x) + 2*sin(x) = sqrt(2)

(2*cos(x) + 2*sin(x))^2 = sqrt(2)^2

4*cos^2(x) + 8*sin(x)*cos(x) + 4*sin^2(x) = 2

4*(cos^2(x) + 2*sin(x)cos(x) + sin^2(x)) = 2

1 + 2*sin(x)cos(x) = 1/2

2*sin(x)*cos(x) = -1/2

sin(2x) = -1/2

Now I need to consider for what values of x is this equation true.

No. What you need is to find the solutions of this equation. Are you
aware of the fact that

sin z = sin w <=> (z = w + 2n pi for some integer _n_) or

I'm aware of the above.


(z = pi - w + 2n pi for some integer _n_)?

And that would be a restricted form of
the first, right? Applying to any
point on the half circle from
0 to pi?


Using this, together with the fact that -1/2 = sin(-pi/6) it is
easy solve the equation.


I don't see the connection. Could you elaborate?

However, this will give also some values which are *not* solutions
of your initial equation. This is so because

2*cos(x) + 2*sin(x) = sqrt(2) (*)

only implies that

(2*cos(x) + 2*sin(x))^2 = sqrt(2)^2,


Could you expound on why the above
only 'implies' but is not equivalent?

but it is not equivalent to it. A better approach would be

(*) <=> sqrt(1/2)*cos(x) + sqrt(1/2)*sin(x) = 1/2

<=> sin(pi/4)*cos(x) + cos(pi/4)*sin(x) = 1/2

<=> sin(pi/4 + x) = 1/2 = sin(pi/6).



How did you arrive at:
sqrt(1/2)*cos(x) + sqrt(1/2)*sin(x) = 1/2

I do the following:
2*cos(x) + 2*sin(x) = sqrt(2)
2(cos(x) + sin(x)) = sqrt(2)
cos(x) + sin(x) = sqrt(2)/2

At which point it is not clear how
you derived a positive 1/2 on the
RHS or how you derived the
sqrt(1/2) on the LHS.

--
conrad

.

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