Question

Number of principal solution(s) of the $\sqrt{sinx}-\frac{1}{\sqrt{sinx}}=cosx$ equation is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$\sqrt{sinx}-\frac{1}{\sqrt{sinx}}=cosx$
$sinx+\frac{1}{sinx}-2={\left(cos\right)}^{2}x=\left(1-{\left(sin\right)}^{2}x\right)$
Let $sinx=t$
$t+\frac{1}{t}-2=1-t^2$
$t^2-2t+1=t\left(1-t\right)\left(1+t\right)$
$\Rightarrow \left(1-t\right)\left[\left(1-t\right)-t\left(1+t\right)\right]=0$
$\Rightarrow \left(1-t\right)\left[1-2t-t^2\right]=0$
$t=1,t^2+2t-1=0$
$sinx=1,sinx=\frac{-2\pm \sqrt{4+4}}{2}$
$sinx=-1\pm \sqrt{2}$
$sinx=1,\Rightarrow x=\frac{\pi }{2}$
also $sinx=\sqrt{2}-1$ and $cosx<0$
$\Rightarrow$ one solution is $\left(0,2\pi \right)$