Answer:
5π/2 it is not a solution for the equation 2sin²x - sinx - 1 = 0
Step-by-step explanation:
∵ 2sin²x - sin x - 1 = 0
* Lets factorize it as a quadratic equation
∴ ( 2sinx + 1)(sinx - 1) = 0
∴ 2sinx + 1 = 0 ⇒ 2sinx = -1 ⇒ sinx = -1/2
* ∵ The value of sinx is -ve
∴ x is in the 3rd or 4th quadrant ⇒ According to ASTC Rule
- ASTC Rule: 1st all +ve , 2nd sin only +ve ,
3rd tan only +ve , 4th cos only +ve
* Let sinα = 1/2 where α is an acute angle
∴ α = π/6
∵ x is in 3rd or 4th quadrant
∴ x = π + α = π + π/6 = 7π/6 or
∴ x = 2π - π/6 = 11π/6
OR
∴ sinx - 1 = 0 ⇒ sinx = 1
∴ x = π/2
∴ ALL values of x are π/2 , 7π/2 , 11π/2 if 0 ≤ x ≤ 2π
∴ 5π/2 it is not a solution for the given equation
