Find all solutions of 4(sin(x)**2)-8cos (x) -8 = 0 in the interval (2pi, 4pi).
(Leave your answers in exact form and enter them as a comma-separated list.)
Using this format, we use
cos^2 x or (cosx)^2 to show the square of cosx
the ^ symbol is used to show exponents.
4 sin^2 x - 8cosx - 8 = 0
4(1 - cos^2 x) - 8cosx - 8 = 0
4 - 4cos^2 x - 8cosx - 8 = 0
4cos^2 x + 8cosx + 4 = 0
cos^2 x + 2cosxs + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
in the first period x = π
so for your domain
x = π + 2π = 3π
To find all solutions of the equation 4(sin(x)^2) - 8cos(x) - 8 = 0 in the interval (2pi, 4pi), we can follow these steps:
Step 1: Simplify the equation.
Using the trigonometric identity sin(x)^2 + cos(x)^2 = 1, we can rewrite the equation as:
4(1 - cos(x)^2) - 8cos(x) - 8 = 0.
Expanding the equation, we have:
4 - 4cos(x)^2 - 8cos(x) - 8 = 0.
Rearranging the terms, we get:
-4cos(x)^2 - 8cos(x) - 4 = 0.
Step 2: Solve the quadratic equation.
Let's consider the quadratic equation in terms of cos(x): -4cos(x)^2 - 8cos(x) - 4 = 0.
To solve the quadratic equation, we can either factorize it or use the quadratic formula.
Factoring the equation may not be straightforward in this case, so we'll use the quadratic formula:
cos(x) = (-b ± sqrt(b^2 - 4ac)) / (2a),
where a = -4, b = -8, and c = -4.
Substituting the values into the formula, we get:
cos(x) = (-(-8) ± sqrt((-8)^2 - 4*(-4)*(-4))) / (2*(-4)),
cos(x) = (8 ± sqrt(64 - 64))/(-8),
cos(x) = (8 ± 0)/(-8).
Simplifying further, we have:
cos(x) = 1 or cos(x) = -1.
Step 3: Find the values of x.
Now that we have cos(x) = 1 and cos(x) = -1, we can find the corresponding values of x using the inverse cosine function (also known as arccosine).
For cos(x) = 1, we have x = arccos(1).
Since the cosine function has a period of 2pi, arccos(1) will give us the solutions in the interval (0, 2pi).
The value of arccos(1) is 0 rad.
For cos(x) = -1, we have x = arccos(-1).
Again, using the concept of the cosine function's periodicity, we need to find all possible values of x in the interval (0, 2pi).
The value of arccos(-1) is pi rad.
Putting it all together, the solutions of the equation 4(sin(x)^2) - 8cos(x) - 8 = 0 in the interval (2pi, 4pi) are:
x = pi, 3pi.
Therefore, the comma-separated exact form of the solutions is:
x = pi, 3pi.