If sqrt(sinx) + cosx = 0, the what is the value of sinx?

Jan 10, 2018

$\sin x = - \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}$.

Explanation:

$\sqrt{\sin} x + \cos x = 0$,

$\Rightarrow \sqrt{\sin} x = - \cos x$,

Squaring, $\sin x = {\cos}^{2} x = 1 - {\sin}^{2} x , \mathmr{and} ,$

${\sin}^{2} x + \sin x = 1$.

Completing Square on the L.H.S., we find,

${\sin}^{2} x + \sin x + \frac{1}{4} = 1 + \frac{1}{4} , i . e . ,$

${\left(\sin x + \frac{1}{2}\right)}^{2} = {\left(\frac{\sqrt{5}}{2}\right)}^{2}$,

$\therefore \sin x + \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$, which gives,

$\sin x = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}$.

But because of $\sqrt{\sin} x , \sin x < 0$ is not admissible.

$\therefore \sin x = - \frac{1}{2} + \frac{\sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}$.

Jan 10, 2018

See below.....

Explanation:

$\sqrt{\sin x} + \cos x = 0$
$\implies \sqrt{\sin} x = - \cos x$
$\implies \sin x = {\cos}^{2} x$
$\implies \sin x = 1 - {\sin}^{2} x$
$\implies {\sin}^{2} x + \sin x - 1 = 0$
$\implies {m}^{2} + m - 1 = 0 \text{ "" where"" } m = \sin x$

Now,
Solution of $\textcolor{red}{m}$ is  This is the answer for $\sin x = \frac{\sqrt{5} - 1}{2}$ as $- 1 < \sin x < + 1$
Graphical representation:-

graph{x^2+x-1 [-7.83, 12.17, -4.81, 5.19]}

Jan 10, 2018

We have to manipulate the equation to a more manageable form, and then solve it.

Explanation:

If $\sqrt{\sin x} = - \cos x$ we can raise this to the square (be careful here, read Note 1 below!)

So we have $\sin x = {\left(- \cos x\right)}^{2} = {\cos}^{2} x$, but ${\sin}^{2} x + {\cos}^{2} x = 1$, so we can use ${\cos}^{2} x = 1 - {\sin}^{2} x$ and we can replace ${\cos}^{2} x$:

$\sin x = 1 - {\sin}^{2} x$ or the same ${\sin}^{2} x + \sin x - 1 = 0$

Now this is a quadratic equation in $\sin x$, we may call $\sin x = y$ and then solve:

${y}^{2} + y - 1 = 0$

We get two solutions:

${y}_{1} = \frac{- 1 + \sqrt{5}}{2}$ and ${y}_{2} = \frac{- 1 - \sqrt{5}}{2}$

But now, we should observe that ${y}_{2} < - 1$ and it can't be the a value for $\sin x$ (see Note 2 below). The other value ${y}_{1} = \frac{- 1 + \sqrt{5}}{2}$ however is a $p o s i t i v e$ number within the range of $\sin x$, so that's a solution for the equation.

Note 1: We have to be careful not to introduce false solutions. Say we have $y = - 1$, we then know that ${y}^{2} = 1$, but when solving this last one we obtain two solutions $y = - 1$ (the original one) and $y = 1$ (a false one). This is because $y = - 1 \implies {y}^{2} = 1$ but not the other way around

Note 2: The solution $y$ of the equation above must be a value within the range $- 1$ and $1$, and be $y \ge 0$ to be possible to take the square root. But $2 < \sqrt{5} < 3$ and so $1 < - 1 + \sqrt{5} < 2$ and then $0 < {y}_{1} = \frac{- 1 + \sqrt{5}}{2} < 1$