IIT JEE Advanced 2007 Paper 1

Since we have
2cos2θ–3sinθ = 0 (1)
we can write as
2sin2θ−(1−2sin2θ) = 0
4sin2θ = 1
sin θ =

1

2

,−

1

2

Now , 2cos2θ−3sinθ = 0 (2)
2 - 2sin2θ−3sinθ = 0
2sin2θ+3sinθ−2 = 0
sin x =

1

2

,−2
⇒ sin θ =

1

2

is common solution.
Therefore, the number of solutions is two:
θ =

π

6

and

5π

6

{where θ ∊ [0, 2π]}