solve the equation -3 cos t = 1 in the interval from 0 to 2p?
-3 cos t = 1
cos(t)=-(1/3)
Look at the plot of cos(t) between 0 and 2π, and note that there are two solutions between π/2 and 3π/2 where cos(t) is negative.
1. function g(x)=-x^2-3x+7 . Find the area under the curve for the domain –4£ x £ 1.
a 71.6 square units
b. 46 square units
c.35.83 square units
d.17.916square units
∫-x^2-3x+7dx between -4 and 1
to evaluate
[-x³/3-(3/2)x²+7x] between -4 and 1.
To solve the equation -3cos(t) = 1 in the interval from 0 to 2π, we can use several steps.
Step 1: Rewrite the equation: cos(t) = -1/3.
Step 2: Use the inverse cosine function (also called arccos or cos^(-1)) to find the values of t.
Since we want the solutions in the interval from 0 to 2π, we need to find the principal value of arccos(-1/3) in that interval.
Step 3: Use a calculator or trigonometric table to find the principal value of arccos(-1/3).
Using a calculator, we find:
arccos(-1/3) ≈ 1.9106 radians.
Since cos function has a periodicity of 2π, we can add any integer multiple of 2π to the principal value to find the other solutions.
Step 4: Find the other solutions in the given interval.
To find the other solutions, we add multiples of 2π to the principal value we found.
t = 1.9106 + 2πk, where k is an integer.
We need to find the values of t that lie in the interval from 0 to 2π.
For k = 0: t = 1.9106.
For k = 1: t = 1.9106 + 2π = 4.0514.
For k = 2: t = 1.9106 + 4π = 7.1922.
...
Keep adding multiples of 2π until we get all the solutions within the given interval.
The solutions for t within the interval from 0 to 2π are:
t = 1.9106, 4.0514, 7.1922, 10.333, ...
Note: It is always a good idea to verify the solutions by plugging them back into the original equation to ensure they satisfy it.