Trigonometry

Question

Solve for 0°<x<180° for the following equation, 8 cosec 2x cot 2x=3

Answer 1

8 cosec 2x . cot 2x = 3

1/sin2x    .   cos2x/sin2x = 3/8

cos2x/sin²2x = 3/8

cos2x = 1 - 2sin²2x

(1- 2sin²2x)/sin²2x = 3/8

1/sin²2x   -    2sin²2x/sin²2x = 3/8

1/sin²2x - 2 = 3/8

1/sin²2x = 3/8  + 2 = 19/8

sin²2x = 8/19

sin 2x = sqrt(8/19) = 0..65

2x = sin^-1(0.65) = 40 degree

x = 20 degree

:)

Answer 2

8 csc(2x) cot(2x) = 3

8 cos(2x)/sin²(2x) = 3

8 cos(2x)/(1 - cos²(2x)) = 3

 

Let u = cos(2x):

 

8 u/(1-u²) =3

8u = 3(1-u²)

3u² + 8u - 3 = 0

 

Solve this quadratic equation, get u = 1/3, -3. Since abs(cos(2x))≤1, u = 1/3 is the only valid solution.

Now solve

cos(2x) = 1/3 ⇒ x = 1/2 cos^-1(1/3) ≈ ± 35°

 

Since 0°<x<180°, x ≈ 35° is the only solution.