Asked by Renee
Solve the equation for solutions in the interval 0<=theta<2pi
Problem 1.
3cot^2-4csc=1
My attempt:
3(cos^2/sin^2)-4/sin=1
3(cos^2/sin^2) - 4sin/sin^2 = 1
3cos^2 -4sin =sin^2
3cos^2-(1-cos^2) =4sin
4cos^2 -1 =4sin
Cos^2 - sin=1/4
(1-sin^2) - sin =1/4
-Sin^2 - sin =-3/4
Sin(sin+1) =3/4
Sin= 3/4. Or sin =-1/4
I don't think these give the right answers for theta. I keep getting an error on my calculator
Problem 1.
3cot^2-4csc=1
My attempt:
3(cos^2/sin^2)-4/sin=1
3(cos^2/sin^2) - 4sin/sin^2 = 1
3cos^2 -4sin =sin^2
3cos^2-(1-cos^2) =4sin
4cos^2 -1 =4sin
Cos^2 - sin=1/4
(1-sin^2) - sin =1/4
-Sin^2 - sin =-3/4
Sin(sin+1) =3/4
Sin= 3/4. Or sin =-1/4
I don't think these give the right answers for theta. I keep getting an error on my calculator
Answers
Answered by
Reiny
from
3cos^2 Ø -4sinØ =sin^2 Ø , why not replace the cos^2 Ø
3(1 - sin^2 Ø) - 4sinØ - sin^2 Ø = 0
..
4sin^2 Ø + 4sinØ - 3 = 0
(2sinØ - 1)(2sinØ + 3) = 0
sinØ = 1/2 or sinØ = -3/2 which is not possible
sinØ = 1/2
Ø = π/6 or 5π/6 , (30° or 150°)
your solution breaks down here:
Sin(sin+1) =3/4
Sin= 3/4. Or sin =-1/4
you can only set factors equal to zero if the right side of your equation is zero.
My second concern is your poor usage of trig function names.
You need an argument after the trig name
e.g. sinØ = .3
sin = .3 is meaningless, and is just that, a sin .
that's like saying √ = 2
3cos^2 Ø -4sinØ =sin^2 Ø , why not replace the cos^2 Ø
3(1 - sin^2 Ø) - 4sinØ - sin^2 Ø = 0
..
4sin^2 Ø + 4sinØ - 3 = 0
(2sinØ - 1)(2sinØ + 3) = 0
sinØ = 1/2 or sinØ = -3/2 which is not possible
sinØ = 1/2
Ø = π/6 or 5π/6 , (30° or 150°)
your solution breaks down here:
Sin(sin+1) =3/4
Sin= 3/4. Or sin =-1/4
you can only set factors equal to zero if the right side of your equation is zero.
My second concern is your poor usage of trig function names.
You need an argument after the trig name
e.g. sinØ = .3
sin = .3 is meaningless, and is just that, a sin .
that's like saying √ = 2
Answered by
Steve
You cannot work this:
sin(sin+1) = 3/4
if sin=3/4, sin+1 = 7/4
This is why we always set things to zero. If a product is zero, one of the factors <b>nust</b> be zero.
sin^2+sin = 3/4
4sin^2+4sin-3=0
(2sin-1)(2sin+3) = 0
sin = 1/2 or sin = -3/2
only sin = 1/2 is allowed, so
θ = π/6, 5π/6
or, since cot^2 = csc^2-1, you have
3(csc^2-1)-4csc - 1 = 0
3csc^2 - 4csc - 4 = 0
(3csc+2)(csc-2) = 0
csc = 2 or -2/3
|csc| is never less than 1, so
csc = 2 is the only solution.
Same as above
sin(sin+1) = 3/4
if sin=3/4, sin+1 = 7/4
This is why we always set things to zero. If a product is zero, one of the factors <b>nust</b> be zero.
sin^2+sin = 3/4
4sin^2+4sin-3=0
(2sin-1)(2sin+3) = 0
sin = 1/2 or sin = -3/2
only sin = 1/2 is allowed, so
θ = π/6, 5π/6
or, since cot^2 = csc^2-1, you have
3(csc^2-1)-4csc - 1 = 0
3csc^2 - 4csc - 4 = 0
(3csc+2)(csc-2) = 0
csc = 2 or -2/3
|csc| is never less than 1, so
csc = 2 is the only solution.
Same as above
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