2sin-3csc=-5

Question

please someone help me find q1 2 3 and 4

Answer 1

The identity    csc = 1 / sin

 

 

2sin(x) - 3(1 / sin(x)) = -5

 

2sin(x) - (3 / sin(x)) = -5

 

 

The LCD is sin(x). 

 

 

(2sin²(x) - 3) / sin(x) = (-5sin(x)) / sin(x)

 

 

Equate numerators.

 

2sin²(x) + 5sin(x) - 3 = 0

 

 

(2sin(x) - 1) (sin(x) + 3) = 0

 

 

2sin(x) - 1 = 0              and              sin(x) + 3 = 0

 

2sin(x) = 1                     and              sin(x) = -3

 

sin(x) = 1/2

 

 

The only solutions are

 

x = 30 degrees

x = 150 degrees         (2nd quadrant)

Answer 2

2sin(x) - 3csc(x) = -5

 

csc(x) = 1/sin(x), so:

 

2sin(x) - 3/sin(x) = -5

2sin²(x) - 3 = -5sin(x)

2sin2(x) + 5sin(x) - 3 = 0

 

Let u = sin(x)

 

2u² + 5u - 3 = 0

 

Factors to:

 

(2u-2)(u+3)

u = sin(x) = 1/2 and -3

 

Since the range of sin(x) is ±1, it cannot equal -3.  Hence sin(x) = 1/2 and x = {30^o, 150^o}

Answer 3

2 sin θ - 3/sin θ = -5

2 sin² θ - 3 = -5 sin θ

2 sin² θ - 3 + 5 sin θ = 0

2 sin² θ + 5 sin θ - 3 = 0

(2 sin θ - 1)(sin θ + 3) = 0

sin θ = 1/2           or        sin θ = -3  (impossible)

 

{π/6, 5π/6}