trignometry please help!!?

Question

theta is represnted by t

this is the question i dont understand hw to slve:

6cos^2t-sint=0

i sub'd cos2t for sin2t

nd got
6-6sin^2t-sint - 5= wt do i do next please explain :(

Answer 1

6cos² t - sin t = 0
6(1 - sin² t) - sin t = 0
6 - 6sin² t - sin t = 0
6 - sin t - 6sin² t = 0
sin t = {1 ± √(1 + 4*6*6)}/(-12) = {-1 ± (√145)}/12
sin t = 0.9201328
the negative value is discarded because it is < -1.

t = arcsin(0.9201328) = 66.945516°
t = 66.945516° + 360n°, 113.05448° + 360n°

Answer 2

Well, you get a second degree equation in sin t. 6 - 6sin^2(t) - sin(t) = 0. (there's no 5). Or 6 sin^2(t) + sin(t) - 6 = 0. So, by Bhaskara, you have 2 solutions, sin(t) = (-1 + Sqrt(1 + 144))/12 and sin(t) = (-1 - Sqrt(1 + 144))/12. If some of this expressions give a number >1 or <-1, then there-s no real t such that sin(t) equal such number. For the values of sin(t) in [-1, 1], youu have to find arc sin (t), thetre are infinitely many solutions, if you dont restrict your domain

Answer 3

You mean the question is
6 cos² t - sin t - 5 = 0

becomes
6 - 6 sin² t - sin t - 5 = 0

Anyway, add the constants
-6 sin² t - sin t + 1 = 0

Multiply -1 to both sides
6 sin² t + sin t - 1 = 0

Then, let
x = sin t.
Therefore,
x² = sin² t

Substitute these to the equation
6x² + x - 1 = 0

Then you can now factor the trinomial
(3x - 1)(x + 1) = 0

We can now sub back x = sin t
(3 sin t - 1)(sin t + 1) = 0

Now,
3 sin t - 1 = 0 or sin t + 1 = 0

We solve for sin t
sin t = 1/3 or sin t = -1

Thus,
t = arcsin 1/3 or t = arcsin -1

For arcsin 1/3, you have irrational solutions, but arcsin -1 = π/2 (4k + 3), where k is an integer.

Therefore, the solutions for t (theta) is
t = arcsin 1/3 or t = ..., -9π/2, -5π/2, -π/2, 3π/2, 7π/2, ...

^_^

Answer 4

Do you mean 6cos2t or 6cos²t?
I am going to assume that you mean the former.
6cos2t - sint = 0
Now cos 2t = cos² t - sin² t = 1 - 2sin² t
6(1 - 2sin² t) - sin t = 0
6 - 12sin² t - sin t = 0
12sin² t + sin t - 6 = 0
(4sin t + 3)(3sin t - 2) = 0
sin t = - 3/4 or sin t = 2/3
t = 229°,311° or 41.9°,138.1°
Hope this is of use to you. Please post the question again if it is not.

Answer 5

Did you mean 6(cos^2)t-sint=0
Then 6(1-sin^2t) - sint = 0
i.e. 6sin^2t+sint-6=0
i.e. sint=[-1+sqrt(145)]/12 and sint=[-1-sqrt(145)]/12
therefore sint=[-1+sqrt(145)]/12 (since -1<=sint<=1)
i.e. sint=0.92
and angel t=66.95 degrees.

Answer 6

Its all goble a duke to me

at a guess i'd say Bob Hoskins ?

Answer 7

6[(cos t)^2]-sin t=0

therefore, 6[1-(sin t)^2]-sin t=0

therefore,6-6(sin t)^2-sin t=0

therefore,6(sin t)^2+sin t -6=0

[sin t-0.9201][sin t+1.08]=0

therefore,sin t=0.9201 or sin t=-1.08

but sin<=1 and sin>= -1

therefore,sin t=.9201

therefore,t=66.9 degrees