Solve (when 0<A<90) : 2cosecA=3sec2A?

Answer 1

Sahsjing and puggy both gave you good answers, but they don't agree. The reason is that they read the problem differently.

Sahsjing read the right side of the equation as:
3 (sec (2A))
And his answer is right if that's what the right side is.

Puggy read the right side as:
3 (sec^2 (A)), which is the same as 3 (sec (A))^2
And his answer is right if that's what the right side is intended to be.

Hope you choose the right answer.
Good luck!

Answer 2

2csc(A) = 3sec^2(A)

Change everything to sines and cosines.

2(1/sin(A)) = 3/(1/cos^2(A))

2/sin(A) = 3/cos^2(A)

2cos^2(A) = 3sinA
2[1 - sin^2(A)] = 3sin(A)

2 - 2sin^2(A) = 3sin(A)

Move everything to the right hand side

0 = 2sin^2(A) + 3sin(A) - 2

And now, this factors like a quadratic, into

0 = (2sin(A) - 1) (sin(A) + 2)

This implies

2sinA - 1 = 0
sin(A) + 2 = 0

We can discard the second equation, because sin(A) is always going to be between -1 and 1. sin(A) + 2 = 0 implies sin(A) = -2, which cannot occur. Therefore our only equation is

2sin(A) - 1 = 0
2sin(A) = 1
sin(A) = 1/2

Since we're given A is in the first quadrant, sin(A) is equal to 1/2 only at the point pi/6. Therefore

A = pi/6

Answer 3

2/sinA = 3/cos2A

Cross multiplication leads to,
3sinA = 2(1-2sin^2 A)

Let x = sinA
4x^2+3x-2 = 0
x = (-3±√41)/8

Since |sinA| ≤ 1,
sinA = x = .4254

A = arcsin(.4254) = 25.18 degrees