Question:
The number of values of x ∈ [0, 2π] satisfying the equation 2 sin2x = 4 + 3 cos x is______________.
Solution:
For x ∈ [0, 2
2sin2x = 4 + 3cos x
i.e 2(1 − cos2x) = 4 + 3cos x
i.e 2 − 2cos2x = 4 + 3cos x
⇒ 2cos2x + 3cos x + 2 = 0
⇒ 2cos2x + 4cos x − cos x + 2 = 0
i.e No solution is possible
∴ No value of x satisfies
2sin2x = 4 + 3cos x