Show that the trigonometric equation has two roots?

Question

Show that the equation sec θ + cosec θ = c has two roots between 0 and 2π if 2 < c^2 <8 ?

Answer 1

squaring both sides .. . .

(1/cosθ)^2 + 2(1/cosθ)(1/sinθ) + (1/sinθ)^2 = c^2

removing the denominators .. . .

sin^2θ + 2sinθcosθ + cos^2θ = c^2 sin^2θ cos^2θ

1 + sin(2θ) = c^2 sin^2(2θ)/4

this means

c^2 sin^2(2θ) - 4sin(2θ) - 4 = 0

solving for the root

sin(2θ) = [4 ± √(16 + 16c^2)] / 2c^2

we also know that

-1 ≤ sin(2θ) ≤ 1

thus

[4 + 4√(c^2+1)]/2c^2 ≤ 1 .. . .. . this is the positive valued

4 √(c^2 + 1) ≤ 2c^2 - 4

16(c^2 + 1) ≤ 4c^4 - 16c^2 + 16

4c^4 - 32c^2 ≥ 0

4c^2 (c^2 - 8) ≥ 0

since c^2 cannot be negative, it turns out c^2 &gt; 8

now, if the root of the quadratic is negative

[4 - 4√(c^2+1)]/2c^2 ≥ -1

-4 √(c^2+ 1) ≥ -2c^2 - 4

2 √(c^2 + 1) ≤ c^2 + 2

4 (c^2  +1) ≤ c^4 + 4c^2 + 4

yielding

c^4 ≥ 0

this happens all the time....

thus in effect, if c^2 ≥ 8, there would be two roots for the equation

sin(2θ) = [4 ± √(16 + 16c^2)] / 2c^2

yielding more than two roots to

sec θ + csc θ = c

if c^2 &lt; 8, the equation

sin(2θ) = [4 ± √(16 + 16c^2)] / 2c^2

is valid only for the negative root

and thus would have only two roots.

... as of now, i cannot find the reason why c^2 &gt; 2 ....