Find all solutions on the interval [0.2pi)
A) -3sin(t)=15cos(t)sin(t)
I have no clue...
b) 8cos^2(t)=3-2cos(t)
All i did was move around the equation to make an quadratic for B.
so
-8cos^2(t)-2cos(t)+3 = 0
what, forgotten your algebra I, now that you're taking trig?
-3sin(t)=15cos(t)sin(t)
15cos(t)sin(t) + 3sin(t) = 0
3sin(t)(5cos(t)+1) = 0
sin(t) = 0
or
cos(t) = -1/5
So, find the 4 values of t which do that.
8cos^2(t)=3-2cos(t)
8cos^2(t)+2cos(t)-3 = 0
(4cos(t)+3)(2cos(t)-1) = 0
cos(t) = -3/4 or 1/2
now go for it.
For part A, let's solve the equation -3sin(t) = 15cos(t)sin(t) on the interval [0, 2π).
To solve this equation, we can start by rearranging it:
-3sin(t) - 15cos(t)sin(t) = 0
Now, we can factor out sin(t) from both terms:
sin(t)(-3 - 15cos(t)) = 0
To find the solutions, we set each factor equal to zero and solve for t:
sin(t) = 0
This equation is true when t is an integer multiple of π.
So, t = 0, π, 2π, ...
Next, let's solve -3 - 15cos(t) = 0 for cos(t):
-3 - 15cos(t) = 0
15cos(t) = -3
cos(t) = -3/15
cos(t) = -1/5
This equation is true when t is in the first and second quadrants.
To find the corresponding angles in the interval [0, 2π), we can use the inverse cosine function:
t = arccos(-1/5)
Now, let's find the values of t in the interval [0, 2π):
t = arccos(-1/5) ≈ 1.8235 radians
However, remember that the previous solution t = 0 was included when solving sin(t) = 0, so we don't need to include it again. Therefore, the solutions on the interval [0.2π) for part A are:
t ≈ 1.8235 rad, π, 2π, ...
For part B, you have correctly rearranged the equation:
-8cos^2(t) - 2cos(t) + 3 = 0
Now, we can solve this quadratic equation. Let's simplify it by multiplying all terms by -1:
8cos^2(t) + 2cos(t) - 3 = 0
We can use the quadratic formula to find the solutions:
cos(t) = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 8, b = 2, and c = -3. Substituting these values into the formula:
cos(t) = (-(2) ± √((2)^2 - 4(8)(-3))) / (2(8))
cos(t) = (-2 ± √(4 + 96)) / 16
cos(t) = (-2 ± √100) / 16
cos(t) = (-2 ± 10) / 16
There are two possible solutions for cos(t):
1) cos(t) = (10 - 2) / 16 = 8/16 = 1/2
2) cos(t) = (-10 - 2) / 16 = -12/16 = -3/4
To find the corresponding angles in the interval [0, 2π), we can use the inverse cosine function:
t = arccos(1/2) and t = arccos(-3/4)
Now, let's find the values of t in the interval [0, 2π):
t₁ = arccos(1/2) ≈ 1.0472 rad
t₂ = arccos(-3/4) ≈ 2.5307 rad
Therefore, the solutions on the interval [0.2π) for part B are:
t ≈ 1.0472 rad, 2.5307 rad