Question

solve the equation csc^2 theta + 3 csc theta - 4 =0 over the interval [ 0,2pi]

Answer 1

kosekant=1/sine, so square=1/sine^2...yu get a thang that go up tu infinity at x=0

& tu - infinity at x=pi (180 deg)

Square it & both gotu + infinity

yer equashun be like x^2+3x-4=0, so hav roots at x=1 & x=-4, so funk=0 at x=1 radian

add 2 pi tu -4 tu get other zero

Answer 2

Given csc^2 theta + 3 csc theta - 4 =0 over the interval [ 0,2pi]

Let theta =t
The mathematical form of the equation is,
csc^2(t) + 3 csc t - 4 = 0

Let u=csc t
u^2 + 3u - 4 = 0
(u + 4)(u - 1) = 0
u + 4 = 0 or u - 1 = 0
u = -4 or u = 1

So, csc t = -4 or 1
1/sin t = -4 or 1
sin t = -1/4 or 1.
t = 2π + arcsin(-1/4),
π - arcsin(-1/4),
or t = π/2

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