NTA Abhyas JEE Main2020MathematicsDeterminantsPractice
If c o s θ - 1 1 c o s 2 θ 4 3 2 7 7 = 0 , then the number of values of θ in 0,2 π is
Options
- A1
- B2
- C3
- D4
Correct answer
B. 2
Step-by-step solution
cos θ 7 + 1 7 cos 2 θ - 6 + 1 7 cos 2 θ - 8 = 0 ⇒ 2 cos 2 θ + cos θ - 2 = 0 ⇒ 4 c o s 2 θ - 2 + cos θ - 2 = 0 ⇒ 4 c o s 2 θ + cos θ - 4 = 0 ⇒ cos θ = - 1 ± 65 8 cos θ = - 1 - 65 8 r e j e c t e d , cos θ = - 1 + 65 8 Hence, 2 solutions are possible