The solution of equations cos2(θ)+sin(θ)+1=0 lies in the interval
-π4,π4
π4,3π4
3π4,5π4
5π4,7π4
Explanation for the correct option
Given: cos2(θ)+sin(θ)+1=0
⇒1-sin2(θ)+sin(θ)+1=0sin2θ+cos2θ=1⇒sin2(θ)-sin(θ)-2=0⇒sin2(θ)-2sin(θ)+sin(θ)-2=0⇒sin(θ)sin(θ)-2+1sin(θ)-2=0⇒sin(θ)+1sin(θ)-2=0
sin(θ)=2 is not possible because the range of sin(θ) has a range of [-1,1].
therefore, sin(θ)+1=0
⇒sin(θ)=-1⇒sin(θ)=sin3π2
comparing both sides
θ=3π2∈5π4,7π4
Hence, option (D) is correct.
A solution of the equation cos2θ+sin θ+1=0,lies in the interval
If the equation cos4θ+sin4θ+λ=0 has real solutions for θ, then λ lies in the interval