Question

If $\cos x + \sqrt {\sin x} = 0$, then x,
A. ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 - 1}}{2}} \right)$
B. $\pi - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 - 1}}{2}} \right)$
C. ${\sin ^{ - 1}}\left( {\dfrac{{1 - \sqrt 5 }}{2}} \right)$
D. \[{\cos ^{ - 1}}\left( {\sqrt {\dfrac{{\sqrt 5 - 1}}{2}} } \right)\]

Answer 1

Hint: We can square the given equation and make it a quadratic equation using trigonometric identities and suitable substitution. Then we can solve the quadratic equation and take sin inverse to get the value of x.

Complete step by step answer:

We have the equation, $\cos x + \sqrt {\sin x} = 0$
For easy calculation, we can subtract both sides with $\sqrt {\sin x} $.
$ \Rightarrow \cos x = - \sqrt {\sin x} $
Now we can take the square on both sides.
$ \Rightarrow {\cos ^2}x = \sin x$
We know that, ${\cos ^2}x = 1 - {\sin ^2}x$. Using this we get,
$1 - {\sin ^2}x = \sin x$
On rearranging we get,
${\sin ^2}x + \sin x - 1 = 0$
Let $u = \sin x$.
$ \Rightarrow {u^2} + u - 1 = 0$
Now we have a quadratic equation. We can solve it using quadratic formula.
\[\therefore u = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting the values of a, b, c, we get,
\[u = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2 \times 1}}\]
\[ \Rightarrow u = \dfrac{{ - 1 \pm \sqrt {1 + 4} }}{2}\]
\[ \Rightarrow u = \dfrac{{ - 1 \pm \sqrt 5 }}{2}\]
$ \Rightarrow \sin x = \dfrac{{ - 1 - \sqrt 5 }}{2},\dfrac{{ - 1 + \sqrt 5 }}{2}$
 The value of $\dfrac{{ - 1 - \sqrt 5 }}{2}$ is less than -1. As it does not belong the range of $\sin x$, we can reject it. So, we have, $\sin x = \dfrac{{\sqrt 5 - 1}}{2}$
Taking the inverse, we get,
$x = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 - 1}}{2}} \right)$
So, the value of x is ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 - 1}}{2}} \right)$
Therefore, the correct answer is option A.

Note: We bring the terms of the equation to either side of the equation for easy squaring. If not, the equation will become complex and the number of terms increases. Then we used trigonometric identities to change the equation in terms of sine function only. For solving the quadratic equation, we used the quadratic formula. We cannot use the factorization method as the roots are not rational numbers. The range of sin function is $\left[ { - 1,1} \right]$. So, the values beyond this interval are not taken as the solution.