Question

A 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

\[\left( 5\pi/4, 7\pi/4 \right)\]
Given:
\[\cos^2 x + \sin x + 1 = 0\]
\[ \Rightarrow (1 - \sin^2 x) + \sin x + 1 = 0\]
\[ \Rightarrow 1 - \sin^2 x + \sin x + 1 = 0\]
\[ \Rightarrow \sin^2 x - \sin x - 2 = 0\]
\[ \Rightarrow \sin^2 x - 2 \sin x + \sin x - 2 = 0\]
\[ \Rightarrow \sin x (\sin x - 2) + 1 (\sin x - 2) = 0\]
\[ \Rightarrow (\sin x - 2) (\sin x + 1) = 0\]

\[\Rightarrow \sin x - 2 = 0\] or \[\sin x + 1 = 0\]
\[\Rightarrow \sin x = 2\] or \[\sin x = - 1\]
\[\sin x = 2\] is not possible.
\[\Rightarrow \sin x = - 1\]
∴ \[\sin x = \sin \frac{3\pi}{2}\]

\[\Rightarrow x = n\pi + ( - 1 )^n \frac{3\pi}{2}, n \in Z\]

The values of x lies in the third and fourth quadrants.
Hence, x lies in \[\left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)\].