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
cosecθ⋅cotθ=2√3
3 answers
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
θ = 30° or 330° or θ = π/6 or 11π/6
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
θ = 30° or 330° or θ = π/6 or 11π/6
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 ....
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 ....
1 answer