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Find all angles, o<Θ<360, that satisfy the equation below, to the nearest 10th of a degree.

4cos^2Θ-2=-1

Find All Angles OltΘlt360 That Satisfy The Equation Below To The Nearest 10th Of A Degree 4cos2Θ21 class=

Sagot :

facundo3141592

Answer:

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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