Trig equation

[pazzy78]

To solve this :

2SinA = Cos A +1

Is there a way to solve without another equation ?

Like the equation 2x + y = 10

Now a solution you can see is x=4, y=2 , but strictly speaking in terms of logic we need another equation to solve this.

Is the Trig equation the same or can we solve it as is using identities.


The answer is Cos A = 3/5 btw, and I can confirm that by subbing, but would be nice to solve properly.

 

[BeansNRice]

[math]2\sin(A)=1+\cos(A)\\ 4\sin^2(A) = \cos^2(A) + 2\cos(A) + 1 \\ 4(1-\cos^2(A)) = \cos^2(A) + 2\cos(A) + 1\\ u=\cos(A) \\ 4(1-u^2) = u^2 + 2u+1 \\ 5u^2+ 2u-3 = 0 \\ u = \dfrac{-2 \pm \sqrt{4+60}}{10} \\ u = \dfrac{-2 \pm 8}{10} = -1,~\dfrac 3 5 \\ A = \cos^{-1}(u) = \pi,~\cos^{-1}\left(\dfrac 3 5\right)[/math]
and of course there are solutions at these values +/- multiples of [imath]2\pi[/imath]

 

[The Highlander]

To solve this :

2SinA = Cos A +1

Is there a way to solve without another equation ?

The answer is Cos A = 3/5 btw, and I can confirm that by subbing, but would be nice to solve properly.

The answer to your question is: No, you cannot "solve" it (in the way you are suggesting) without further information, for example, a range, eg: 0 < A < 1.

The answer is not "Cos A = 3/5 btw", that is only one of an infinite number of possible answers as @BeansNRice pointed out at the end of his post.

Just as in your example of x = 4 & y = 2 being a solution of the equation: 2x + y = 10, further information is needed to arrive at the correct values for x & y (x = 6 & y = ˉ2 or x = ˉ1 & y = 12 would fit the equation equally well!), if , indeed, x & y are unknowns and not just variables.

The trigonometric equation you quote also produces an infinite number of solutions for A unless and until further information is provided.

Hope that helps.

 

[pazzy78]

[math]2\sin(A)=1+\cos(A)\\ 4\sin^2(A) = \cos^2(A) + 2\cos(A) + 1 \\ 4(1-\cos^2(A)) = \cos^2(A) + 2\cos(A) + 1\\ u=\cos(A) \\ 4(1-u^2) = u^2 + 2u+1 \\ 5u^2+ 2u-3 = 0 \\ u = \dfrac{-2 \pm \sqrt{4+60}}{10} \\ u = \dfrac{-2 \pm 8}{10} = -1,~\dfrac 3 5 \\ A = \cos^{-1}(u) = \pi,~\cos^{-1}\left(\dfrac 3 5\right)[/math]
and of course there are solutions at these values +/- multiples of [imath]2\pi[/imath]

Beautiful solution !
I had squared both sides but I abandoned it thinking it would be too messy !

I was even thinking of the (1 - cos²A) = sin²A but with the 4sin²A thought it would get too complex... bloody **** like I can't factor out the 4....

Anyway this was part of a larger problem , maybe later I'll post the whole q...

Thanks again!