So I posted these...but no one answered them. Are they right?
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 check whether your solutions are correct, let's go through the process of solving these equations.
1. Solve 4sin^2 x + 4√2 cos x - 6 = 0 for all real values of x:
To solve this quadratic equation, we can use the quadratic formula. First, let's rewrite the equation in terms of cos(x):
4(1 - cos^2 x) + 4√2 cos x - 6 = 0
4 - 4cos^2 x + 4√2 cos x - 6 = 0
-4cos^2 x + 4√2 cos x - 2 = 0
Now we can apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = -4, b = 4√2, and c = -2. Substituting these values into the quadratic formula, we get:
x = (-4√2 ± √((4√2)^2 - 4(-4)(-2))) / (2(-4))
x = (-4√2 ± √(32 + 32)) / -8
x = (-4√2 ± √64) / -8
x = (-4√2 ± 8) / -8
Simplifying further:
x = (-√2 ± 2) / -2
x = √2/2 ± 1
Therefore, the possible solutions for x are:
x = π/4 + 2πk where k is an integer
x = 7π/4 + 2πk where k is an integer
Your answer of (π/4) + 2πk and (7π/4) + 2πk is correct.
2. Solve 2cos^2 x - 5cos x + 2 = 0 for the principal values of x:
We can solve this quadratic equation by factoring. The equation can be factored as:
(2cos x - 1)(cos x - 2) = 0
To find the values of cos x that make the equation true, we set each factor equal to zero:
(2cos x - 1) = 0 or (cos x - 2) = 0
Solving the first factor:
2cos x - 1 = 0
2cos x = 1
cos x = 1/2
Taking the inverse cosine (cos^(-1)) of both sides:
x = cos^(-1)(1/2)
The principal value of cos^(-1)(1/2) is 60 degrees.
Solving the second factor:
cos x - 2 = 0
cos x = 2
However, the cosine function has a range of -1 to 1, so there are no real values of x that satisfy this equation. Therefore, there is only one principal value solution which is x = 60 degrees.
Hence, your answer of 60 and 300 degrees is correct.