Question

The solution of the equation cos^2 theta+ sin theta +1 = 0 lies in
the interval.​

Answer 1

Step-by-step explanation:

We have,

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

$\sf{ \implies \: 1 - sin^{2} ( \theta) + sin( \theta) + 1 = 0 }$

$\sf{ \implies \: - sin^{2} ( \theta) + sin( \theta) + 2 = 0 }$

$\sf{ \implies \: sin^{2} ( \theta) - sin( \theta) - 2 = 0 }$

$\sf{ \implies \: sin^{2} ( \theta) - 2 sin( \theta) + sin (\theta) - 2 = 0 }$

$\sf{ \implies \: \sin( \theta) \{ sin( \theta) - 2 \}+1 \{ sin (\theta) - 2 \}= 0 }$

$\sf{ \implies \: \{ \sin( \theta) +1 \} \{ sin (\theta) - 2 \}= 0 }$

$\sf{ \implies \: \sin( \theta) +1 = 0 \: \: or \: \: sin (\theta) - 2 = 0 }$

$\sf{ \implies \: \sin( \theta) = - 1 \: \: or \: \: sin (\theta) = 2 }$

But $sin(x) \in[-1,1]$, so,

$\sf{ \implies \: \sin( \theta) = - 1 }$

$\sf{ \implies \: \theta= (4n + 3) \frac{\pi}{2} \: \: \: \: \: \:, \forall \: n \in \mathbb{Z} }$