The number of values of x ∈ [0, 2π]
Question:

The number of values of x ∈ [0, 2π] satisfying the equation 2 sin2x = 4 + 3 cos x is______________.

Solution:

For ∈ [0, 2ππ]

2sin2= 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

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