The solution of the equation cos 2 θ+ sin θ+1=0, lies in the interval

Question

The solution of the equation cos 2 θ+ sin θ+1=0, lies in the interval (A) (-(π/4), (π/4)) (B) ((π/4), (3 π/4)) (C) ((3 π/4), (5 π/4)) (D) ((5 π/4), (7 π/4))

Tardigrade Admin · Accepted Answer

We have sin 2 θ- sin θ-2=0 ⇒ ( sin θ+1)( sin θ-2)=0 As sin θ ≠ 2 ∴ sin θ=-1= sin (3 π/2) ∴ θ=(3 π/2)=(6 π/4) ∈((5 π/4), (7 π/4))

Q. The solution of the equation $cos^{2} θ + sin θ + 1 = 0$ , lies in the interval

$(- \frac{π}{4}, \frac{π}{4})$

$(\frac{π}{4}, \frac{3 π}{4})$

$(\frac{3 π}{4}, \frac{5 π}{4})$

$(\frac{5 π}{4}, \frac{7 π}{4})$

We have $sin^{2} θ - sin θ - 2 = 0$
$\Rightarrow (sin θ + 1) (sin θ - 2) = 0$
As $sin θ \neq = 2$
$∴ sin θ = - 1 = sin \frac{3 π}{2}$
$∴ θ = \frac{3 π}{2} = \frac{6 π}{4} \in (\frac{5 π}{4}, \frac{7 π}{4})$

Q. The solution of the equation cos2θ+sinθ+1=0, lies in the interval

(−4π​,4π​)

(4π​,43π​)

(43π​,45π​)

(45π​,47π​)