Solve the equation for the interval [0, 2π).
2 sin2 x + sin x = 1
SOLUTION :
2 sin^2 (x) + sin (x) = 1 (x in the interval [0, 2π) )
=> 2 sin^2 (x) + sin (x) - 1 = 0
=> (2 sin (x) - 1) (sin (x) + 1) = 0
=> sin(x) = 1/2 ; sin (x) = - 1
=> x = π/6, 5π,/6 ; 3π/2 (in the interval [0, 2π) )
These values do satisfy the given equation.
So,
x = {π/6, 5π/6, 3pi/2} ( in the interval [0, 2π) ) : 4th option (ANSWER).
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