Exams Joint Entrance Examination Mathematics Trigonometric Functions and Equations Question 30 Question 30 of 33 Medium MCQ
If θ ∈ [ − 2 π , 2 π ] \theta \in[-2 \pi, 2 \pi] θ ∈ [ − 2 π , 2 π ] , then the number of solutions of 2 2 cos 2 θ + ( 2 − 6 ) cos θ − 3 = 0 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 2 2 cos 2 θ + ( 2 − 6 ) cos θ − 3 = 0 , is equal to:
Solutions · 1 Post your solution Official solution
\[ 2 2 cos 2 θ + ( 2 − 6 ) cos θ − 3 = 0 2 2 cos 2 θ + 2 cos θ − 6 cos θ − 3 = 0 ( 2 cos θ − 3 ) ( 2 cos θ + 1 ) = 0 ⇒ cos θ = 3 2 or cos θ = − 1 2 θ = { − 11 π 6 , − 5 π 4 , − 3 π 4 , − π 6 , π 6 , 3 π 4 , 5 π 4 , 11 π 6 } ⇒ 8 (solution) \begin{aligned}
& 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 \\
& 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\
& (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\
& \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text { or } \cos \theta=\frac{-1}{\sqrt{2}} \\
& \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\
& \Rightarrow 8 \text { (solution) }
\end{aligned} 2 2 cos 2 θ + ( 2 − 6 ) cos θ − 3 = 0 2 2 cos 2 θ + 2 cos θ − 6 cos θ − 3 = 0 ( 2 cos θ − 3 ) ( 2 cos θ + 1 ) = 0 ⇒ cos θ = 2 3 or cos θ = 2 − 1 θ = { 6 − 11 π , 4 − 5 π , 4 − 3 π , 6 − π , 6 π , 4 3 π , 4 5 π , 6 11 π } ⇒ 8 (solution) \] Answer the question to unlock