Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x.
My answer: (pi/4)+2(pi)k, (7pi/4) + 2(pi)k
Solve 2cos^2 x-5cosx+2=0 for principal values of x.
My answer: 60 and 300 degrees.
To solve the equation 4sin^2 x + 4sqrt(2) cos x - 6 = 0, we can start by factoring.
First, let's factor out the common factor of 2 from the first two terms:
2(2sin^2 x + sqrt(2) cos x) - 6 = 0
Next, we can rewrite the equation using the identity sin^2 x + cos^2 x = 1:
2(2sin^2 x + 2cos^2 x - cos^2 x + sqrt(2) cos x) - 6 = 0
Simplifying further, we have:
2(4cos^2 x - sqrt(2) cos x + 1) - 6 = 0
Now, let's divide the entire equation by 2:
4cos^2 x - sqrt(2) cos x + 1 - 3 = 0
Rearranging the terms, we get:
4cos^2 x - sqrt(2) cos x - 2 = 0
Now, we can solve for cos x using the quadratic formula:
cos x = [sqrt(2) ± √(sqrt(2)^2 - 4(4)(-2))] / (2(4))
cos x = [sqrt(2) ± √(2 + 32)] / 8
cos x = [sqrt(2) ± √(34)] / 8
To find the values of x, we need to find the inverse cosine of each value of cos x. Let's solve for the first case:
cos x = [sqrt(2) + √(34)] / 8
x = cos^(-1)[(sqrt(2) + √(34)) / 8]
x = cos^(-1)(sqrt(2) + √(34)) / 8
Next, let's solve for the second case:
cos x = [sqrt(2) - √(34)] / 8
x = cos^(-1)[(sqrt(2) - √(34)) / 8]
x = cos^(-1)(sqrt(2) - √(34)) / 8
Therefore, the solutions for the equation 4sin^2 x + 4sqrt(2) cos x - 6 = 0 are:
x = cos^(-1)(sqrt(2) + √(34)) / 8 + 2πk
x = cos^(-1)(sqrt(2) - √(34)) / 8 + 2πk
where k is an integer.
Now, let's solve the equation 2cos^2 x - 5cos x + 2 = 0 for principal values of x.
To start, we can solve this quadratic equation by factoring:
(2cos x - 1)(cos x - 2) = 0
Setting each factor equal to zero, we have two cases:
2cos x - 1 = 0 or cos x - 2 = 0
For the first case, solving for cos x:
2cos x = 1
cos x = 1/2
x = cos^(-1)(1/2)
x = π/3
For the second case, solving for cos x:
cos x = 2
No real solutions exist for this case, as the range of cosine values is -1 to 1.
Therefore, the solution for the equation 2cos^2 x - 5cos x + 2 = 0 for principal values of x is:
x = π/3 (or 60 degrees)
To solve the equation 4sin^2(x) + 4√2cos(x) - 6 = 0 for all real values of x, we can use the quadratic formula. Let's start by rearranging the equation:
4sin^2(x) + 4√2cos(x) - 6 = 0
We can rewrite this equation in terms of the trigonometric identity sin^2(x) + cos^2(x) = 1 by substituting 1 - cos^2(x) for sin^2(x):
4(1 - cos^2(x)) + 4√2cos(x) - 6 = 0
Expanding and rearranging, we get:
4cos^2(x) + 4√2cos(x) - 2 = 0
Now, let's solve this quadratic equation for cos(x). We can use the quadratic formula:
cos(x) = [-b ± sqrt(b^2 - 4ac)] / (2a)
In this case, a = 4, b = 4√2, and c = -2. Plugging in these values, we have:
cos(x) = [-(4√2) ± sqrt((4√2)^2 - 4(4)(-2))] / (2(4))
Simplifying further:
cos(x) = [-4√2 ± sqrt(32 + 32)] / 8
cos(x) = [-4√2 ± sqrt(64)] / 8
cos(x) = [-4√2 ± 8] / 8
Now, we can simplify these expressions:
cos(x) = (-√2 ± 2) / 2
To find the values of x, we need to find the values of cos(x) that satisfy these expressions. Let's consider each case separately.
Case 1: cos(x) = (-√2 + 2) / 2
To find the values of x for this case, we need to find the inverse cosine of (-√2 + 2) / 2. Using a calculator, we find:
x = arccos((-√2 + 2) / 2)
This gives us one solution for x.
Case 2: cos(x) = (-√2 - 2) / 2
Again, we need to find the inverse cosine to find the values of x. Using a calculator:
x = arccos((-√2 - 2) / 2)
This gives us another solution for x.
In both cases, x can be written as:
x = arccos((-√2 + 2) / 2) + 2πk
x = arccos((-√2 - 2) / 2) + 2πk
where k is an integer. Therefore, the solutions for all real values of x are:
x = arccos((-√2 + 2) / 2) + 2πk
x = arccos((-√2 - 2) / 2) + 2πk
where k is an integer.
Now let's solve the equation 2cos^2(x) - 5cos(x) + 2 = 0 for the principal values of x.
We can factor this quadratic equation to make it easier to solve:
(2cos(x) - 1)(cos(x) - 2) = 0
Now we set each factor equal to zero and solve for cos(x):
1) 2cos(x) - 1 = 0
Adding 1 to both sides, we get:
2cos(x) = 1
Dividing both sides by 2, we obtain:
cos(x) = 1/2
Taking the inverse cosine of 1/2, we find:
x = arccos(1/2)
2) cos(x) - 2 = 0
Adding 2 to both sides:
cos(x) = 2
However, the cosine function only takes values between -1 and 1, so there are no real solutions for this equation.
Therefore, the only real solution for the equation 2cos^2(x) - 5cos(x) + 2 = 0 is:
x = arccos(1/2)
Now, let's find the principal values of x. The inverse cosine function has a range of [0, π], so the principal values of x will be within this range.
Using a calculator, we find:
x ≈ 60° and x ≈ 300°
Therefore, the principal values of x for the equation 2cos^2(x) - 5cos(x) + 2 = 0 are:
x = 60° and x = 300°.