Determine the general solution, in radians, for 2sin B = 3tan^2 B by solving algebraically

Answer 1

2sinx = 3tan^2(x) = 3sin^2(x)/cos^2(x)

2 = 3sinx/cos^2(x)= 3sinx/(1-sin^2(x))

2-2sin^2(x) = 3sinx

let y = sinx

2-2y^2 = 3y

2y^2 +3y -2 = 0

(2y -1)(y+2) = 0

y = 1/2,-2 = sinx

x = sin^-1(1/2), sin^-1(0) = 0 and 30 degrees+ 360n, or 0

and pi/6+ 2pin or 5pi/6 + 2pin, where n=any integer

2sin0 = 3tan^2(0)

2(0)= 3(0^2)

0=0, x=0 is one solution

2sin30 = 3tan^2(30)

2(1/2) =3(1/sqr3)^2 = 3(1/3)

1 = 1, x= 30 degrees is another solution

2sin150 = 3tan^2(150)

2(1/2) = 3(1/-sqr3)^2

1 = 3/3 = 1

x= 150 is another solution

each plus 360n where n= any integer

Answer 2

2sin x = 3tan²x means that

So sin x(2 – 2sin²x) = sin x(3sin x) which means that

sin x(2sin²x + 3sin x – 2) = sin x(2sin x – 1)(sin x + 2) = 0

sin x = 0 leads to one group of solutions

The quadratic solutions are sin x = 1/2 and sin x = –2

The second solution is not possible because 1 ≥ sin x ≥ –1

So we have sin x = ½ leading to a second group of solutions.

The solutions on [0,2π) are x =0, π/6 , 5π/6, and π

General solutions are x = πk, π/6 + 2πk and 5π/6 + 2πk where k is any integer.