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 =  , ,  .

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

72 is the answer to this problem