Question

cosecθ⋅cotθ=2√​3
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Answer 1

cscθ cotθ = 2√3
checking the easy stuff first, you know that
csc(π/6) = 2 and cot(π/6) = √3
so, clearly θ = π/6 is one solution
θ = -π/6 also works, since there both csc and cot are negative

Or, if you want to go the algebraic route,
cosθ/sin^2θ = 2√3
cosθ = 2√3 (1-cos^2θ)
2√3 cos^2θ + cosθ - 2√3 = 0
cosθ = (-1±√(1+48))/(4√3) = (-1±7)/(4√3)
cosθ = √3/2 or -2/√3
θ = π/6

Answer 2

I suppose we are solving for θ

cosecθ⋅cotθ=2√​3
(1/sinθ)(cos/sin) = 2√3
cosθ/sin^2θ = 2√3
2√3 sin^2 θ = cosθ
2√3(1-cos^2 θ) = cosθ
2√3cos^2 θ + cosθ - 2√3 = 0

cosθ = (-1 ± √(1 - 4(2√3)(2√3))/(4√3)
= (-1 ± √49)/(4√3)
= 3/(2√3) or - 2/√3

if cosθ = 3/(2√3), θ = 30° or 330°
if cosθ = -2/√3 <----- not possible since -1 < cosθ < 1

<b>θ = 30° or 330°</b> or <b>θ = π/6 or 11π/6</b>

Answer 3

csc θ ⋅ cot θ = 2√​3

Subtract 2√​3 to both sides

csc θ ⋅ cot θ - 2√​3 = 0

1 / sin θ ⋅ cos θ / sin θ - 2√​3 = 0

cos θ / sin² θ - 2√​3 = 0

cos θ / sin² θ - 2√​3 ⋅ sin² θ / sin² θ = 0

( cos θ - 2√​3 ⋅ sin² θ ) / sin² θ = 0

Fraction = 0 when numerator = 0

So you must solve:

cos θ - 2√​3 ⋅ sin² θ = 0

Use identity:

sin² θ = 1 - cos² θ

cos θ - 2√​3 ⋅ ( 1 - cos² θ ) = 0

cos θ - 2√​3 + 2√​3 ⋅ cos² θ = 0

2√​3 ⋅ cos² θ + cos θ - 2√​3 + = 0

Use substitution:

cos θ = x

2√​3 ⋅ x² + x - 2√​3 = 0

Use quadratic formula:

x1/2= [ - b ± √ ( b² - 4 ⋅ a ⋅ c ) ] / 2 ⋅ a

In this case:

a = 2√​3 , b = 1 , c = - 2√​3

x1/2= [ - 1 ± √ 1² - 4 ⋅ 2 √​3 ⋅ ( - 2√​3 ) ] / 2 ⋅ 2√​3 =

[ - 1 ± √ ( 1 + 8 √​3 ⋅ 2√​3 ) ] / 4√​3 =

[ - 1 ± √ ( 1 + 16 √​3 ⋅√​3 ) ] / 4√​3 =

[ - 1 ± √ ( 1 + 16 ⋅ 3 ) ] / 4√​3 =

[ - 1 ± √ ( 1 + 48 ) ] / 4√​3 =

( - 1 ± √ 49 ) / 4√​3 =

( - 1 ± 7 ) / 4√​3

The solutions are:

x = ( - 1 - 7 ) / 4√​3 = - 8 / 4√​3 = - 4 ⋅ 2 / 4√​3 = - 2 / √​3

and

x = ( - 1 + 7 ) / 4√​3 = 6 / 4√​3 = 2 ⋅ 3 / 2 ⋅ 2 ⋅ √​3 = 3 / 2 √​3 = √​3 ⋅ √​3 / 2 √​3 = √3 / 2

Substitute back :

cos θ = x

The solutions are:

cos θ = - 2 √3 / 3 and cos θ = √3 / 2

Solution cos θ = - 2 √3 / 3 = -1.1547005 you must discard because cosine can be - 1 ÷ 1.

So solution is:

cos θ = √3 / 2

In interval 0 to 2 π for angles θ = π / 6 and θ = 11 π / 6 , cos θ = √3 / 2

General solutions for cos θ = √3 / 2 :

θ = π / 6 ± 2 π n

and

θ = 11 π / 6 ± 2 π n

Because period of cos θ = 2 π

n = 1 , 2 , 3 ....