Find all angles, o<Θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4cos^2Θ-2=-1

30° and 150°

Step-by-step explanation:

We want to find the angles in the range:

0° < θ < 360°

Such that:

4*cos^2(θ) - 2 = 1

The first part is to isolate the variable θ in one side of the equation, let's do that:

4*cos^2(θ) = 1 + 2 = 3

cos^2(θ) = 3/4

cos(θ) = ± √(3/4) = ± (√3)/2

Then we have two possible values of θ, one for:

cos(θ) =  (√3)/2

And another for

cos(θ) = - (√3)/2

For both cases, we can use the inverse cosine function, Acos(x)

that has the property Acos( cos(x) ) = x

Then:

cos(θ) =  (√3)/2

Acos( cos(θ) ) = Acos(  (√3)/2)

θ = Acos(  (√3)/2) = 30°

And for the other case we have:

θ = Acos( - (√3)/2) = 150°

Then the two values of theta are: 30° and 150°

(We can see that both of these values are in the range 0° < θ < 360°)

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explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer

step-by-step explanation: