Question

Stumped?...
Supposed to factor or use quadratic to find solutions but I can't get this to come out as factorable?
Sqrt3 sin x - sqrt1+sin^2 x =0
I get 3sint - 1 +sin^2t =0 "after squaring"
but Im not understanding why i cant factor?

Answer 1

because your equation is not an algebraic equation

Answer 2

Can you write this using the equation tool please? I'm struggling to understand what you've written down - in particular, the sqrt1 part which seems like you're missing parentheses around something.

Answer 3

I'm guessing you mean
\[\sqrt{3\sin x} - \sqrt{1 + \sin^{2}x} = 0\] If so, we can do
\[\sqrt{3sinx} = \sqrt{1+\sin^{2}x}\]
\[(\sqrt{3sinx})^{2} =(\sqrt{1+\sin^{2}x})^{2} \implies 3sinx = 1+ \sin^{2}x\] From there we can set everything equal to 0 and treat what we have like a quadratic
\[\sin^{2}x - 3sinx + 1 = 0\] Now this doesn't factor cleanly, but since it is quadratic in sine, we can use quadratic formula on our equation. Doing so gives us:
\[\frac{ 3 \pm \sqrt{3^{2} - 4(1)(1)} }{ 2 } \implies \frac{ 3 \pm \sqrt{5} }{ 2 }\] Therefore we have the two equations:
\[sinx = \frac{ 3 + \sqrt{5} }{ 2 }\]
\[sinx = \frac{ 3 - \sqrt{5} }{ 2 }\]
Now sinx has a range of [-1,1], meaning there is no solution to any equation where sinx is equal to a number below -1 or above 1. Since \(\frac{3+\sqrt{5}}{2}\) is greater than 1, the equation \(sinx = \frac{ 3 + \sqrt{5} }{ 2 }\) has no solution. Since \(-1 \le\ \frac{3-\sqrt{5}}{2} \le\ 1\) we can find a solution to \(sinx = \frac{ 3 - \sqrt{5} }{ 2 }\) by taking the inverse sine of both sides, giving us x = \(sin^{-1}(\frac{ 3 - \sqrt{5} }{ 2 })\). If you need an approximation, you can get your calculator and come up with \(x\ \approx\ .3919\) radians or \(x\approx 22.46 \) degrees