Find all solutions in the interval [0, 2π).

4 sin2 x - 4 sin x + 1 = 0

ANSWER

EXPLANATION

The given trigonometric equation is

This is a quadratic equation in sinx.

We split the middle term to obtain,

Factor by grouping to get,

This implies that,

This gives us,

in the first quadrant.

Or

in the second quadrant.

x = , ,  .

Step-by-step explanation:

The given equation is 4 sin²x - 4 sin x + 1 = 0

(2sinx)² - 2(2sinx) + 1 = 0

(2sinx - 1 )² = 0

Sinx = ⇒ x = sin⁻¹ ( )

So between the interval [0, 2π] value of x will be  ,  ,  

[Since sine is positive in 1st 3rd and 4th quadrant]

So value of x will be x =  , ,  .

step-by-step explanation:

47

step-by-step explanation: