Respuesta :

Ashraf82

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

Blacklash

Answer:

5pi/2 is a solution for the equation 2sin²x -sin x -1=0

Step-by-step explanation:

We need to check 5pi/2 is a solution for the equation 2sin²x -sin x -1=0.

Substituting 5pi/2 in equation

       [tex]2sin^2x-sin x-1=2sin^2\left (\frac{5\pi}{2} \right )-sin\left (\frac{5\pi}{2} \right )-1\\\\=2sin^2\left (2\pi +\frac{\pi}{2} \right )-sin\left (2\pi +\frac{\pi}{2} \right )-1\\\\=2sin^2\left (\frac{\pi}{2} \right )-sin\left (\frac{\pi}{2} \right )-1\\\\=2\times 1-1-1=0[/tex]

So 5pi/2 is a solution for the equation 2sin²x -sin x -1=0.

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