Question

The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval

Options

• \[\left( - \pi/4, \pi/4 \right)\]

   

• \[\left(\pi/4,3 \pi/4 \right)\]

   

• \[\left( 3\pi/4, 5\pi/4 \right)\]

   

• \[\left( 5\pi/4, 7\pi/4 \right)\]

   

Answer 1

Given equation:
\[\cos^2 x + \sin x + 1 = 0\]
\[ \Rightarrow (1 - \sin^2 x) + \sin x + 1 = 0\]
\[ \Rightarrow 2 - \sin^2 x + \sin x = 0\]
\[ \Rightarrow \sin^2 x - \sin x - 2 = 0\]
\[ \Rightarrow \sin^2 x - 2 \sin x + \sin x - 2 = 0\]
\[ \Rightarrow \sin x ( \sin x - 2 ) + 1 ( \sin x - 2 ) = 0\]
\[ \Rightarrow (\sin x - 2) ( \sin x + 1) = 0\]
\[\Rightarrow \sin x - 2 = 0\] or \[\sin x + 1 = 0\]
\[\Rightarrow \sin x = 2\] or sin x = - 1
Now,

sin x = 2 is not possible
And,
\[\sin x = - 1 \]
\[ \Rightarrow \sin x = \sin \frac{3\pi}{2} \]
\[ \Rightarrow x = n\pi + \left( - 1 \right)^n \frac{3\pi}{2}\]
For n = 0,
\[x = \frac{3\pi}{2}\], for n = 1,
\[x = \frac{7\pi}{2}\] and so on.
Hence,
\[\frac{3\pi}{2}\] lies in the interval 

\[\left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)\].