Question

Solve ${\cos }^2\theta +\sin \theta +1=0$

Answer 1

${\cos }^2\theta +\sin \theta +1=0$

$1-{\sin }^2\theta +\sin \theta +1=0$ (${\cos }^2\theta =1-{\sin }^2\theta$)

${\sin }^2\theta -\sin \theta -2=0$

${\sin }^2\theta -2\sin \theta +\sin \theta -2=0$

$\sin \theta \left(\sin \theta -2\right)+1\left(\sin \theta -2\right)=0$

$\left(\sin \theta +1\right)\left(\sin \theta -2\right)=0$

$\sin \theta +1=0\Rightarrow \sin \theta =-1\Rightarrow \sin \theta =\sin \frac{3\mathrm{\pi }}{2}\Rightarrow \theta =\frac{3\mathrm{\pi }}{2}$

$\sin \theta =2isnotpossible\sin cetherangeof\sin is\left[-1,1\right]$