Question

The solution of the equation ${\cos }^{2}x+\sin x+1=0,x\in [0,2\pi ]$ lies in the interval :

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

Answer 1

The correct option is D $(\frac{5\pi }{4},\frac{7\pi }{4})$

${\cos }^{2}x+\sin x+1=0\Rightarrow 1-{\sin }^{2}x+\sin x+1=0\Rightarrow (\sin x+1)(\sin x-2)=0\Rightarrow \sin x=-1=\sin \frac{3\pi }{2}\Rightarrow x=(4n-1)\frac{\pi }{2}\Rightarrow x=\frac{3\pi }{2}\in (\frac{5\pi }{4},\frac{7\pi }{4})$