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# Trigonometry

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

### 2 Answers by Expert Tutors

Amarjeet K. | Professional Engineer for Math and Science TuroringProfessional Engineer for Math and Scien...
4.6 4.6 (8 lesson ratings) (8)
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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

:)

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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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.