Question

$\text{A solution of the equation}co{s}^2\theta +sin\theta +1=0,liesintheinterval$

• A. $(-\pi /4,\pi /4)$
• B. $(\pi /4,3\pi /4)$
• C. $(3\pi /4,5\pi /4)$
• D. $(5\pi /4,7\pi /4)$

Answer 1

The correct option is D

$(5\pi /4,7\pi /4)$

$Given:co{s}^2\theta +sin\theta +1=0\Rightarrow (1-si{n}^2\theta )+sin\theta +1=0\Rightarrow 1-si{n}^2\theta +sin\theta +1=0\Rightarrow si{n}^2\theta -sin\theta -2=0\Rightarrow {\sin }^2\theta -2sin\theta +sin\theta -2=0\Rightarrow sin\theta (sin\theta -2)+1(sin\theta -2)=0\Rightarrow (sin\theta -2)(sin\theta +1)=0\Rightarrow sin\theta -2=0orsin\theta +1=0\Rightarrow sin\theta =2orsin\theta =-1\Rightarrow sin\theta =2isnotpossible.\Rightarrow sin\theta =-1\therefore sin\theta =sin\frac{3\pi }{2}\Rightarrow \theta =n\pi +(-1{)}^n\frac{3\pi }{2},n\in Z\text{The values of q lies in the third and fourth quadrants.}Hence,\theta liesin(\frac{5\pi }{4},\frac{7\pi }{4}).$