Question

Solve on the interval (0,2%): (sin x + 1)2 sin x-3 sinx-2) = 0 O A. x = 7T,X = 2 x=51 O 3 7T O В. x= N ***= 7+ x = 117 6 6 T T o C. X=2.7 x= 3 O D. x=2x-x-5 2 * = 4

Answer 1

Given:

(sinx + 1)(2sin²x - 3sinx - 2) = 0

sinx + 1 = 0

sinx = -1

Take the arcsin of both-side

x = sin⁻' (-1)

x = -90= -90 + 360 = 270 = 3π / 2

x =3π / 2

Similarly,

2sin²x - 3sinx - 2 = 0

let y = sinx

2y² - 3y - 2 = 0

Factorising the above;

2y² - 4y + y - 2 = 0

2y(y-2) + 1(y-2) = 0

(2y+1)(y-2) = 0

Put sin x back in for y so we can see what it's really supposed to be.

(2sinx + 1 )(sinx -2) = 0

Either one of the terms 2sin x + 1 or sin x - 2 can be 0 for this equation to be true. Let's solve each of them separately.

2sinx + 1 = 0

2sinx = -1

sinx = -1/2

Take the arcsin of both-side.

x= sin⁻' (-1/2)

x = - 30 = -30+360 = 330 ,

x = 7π /6 or 11π/6

Therefore, the correct option is B. x= 3π / 2 , 7π /6 , 11π/6