Question

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

Answer 1

5π/2 it is not a solution for the equation 2sin²x - sinx - 1 = 0

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

Answer 2

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

Explanation:

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

Substituting 5pi/2 in equation

`2sin^2x-sin x-1=2sin^2\left ((5\pi)/(2) \right )-sin\left ((5\pi)/(2) \right )-1\\\\=2sin^2\left (2\pi +(\pi)/(2) \right )-sin\left (2\pi +(\pi)/(2) \right )-1\\\\=2sin^2\left ((\pi)/(2) \right )-sin\left ((\pi)/(2) \right )-1\\\\=2* 1-1-1=0`

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