solve 4 sin^2x + 4 sqrt 2 cos x-6=0 for all real values of x.
Going by the example in the book I got to: 4 cos^2x + 4sqrt2 cosx -10 = 0 but do not know how to proceed.
Any help would be great.
Thanks
if you replaced sin^2x with 1 - cos^2x and simplified correctly you should have had
4cos^2x - 4√2cosx + 2 = 0
solve this as a quadratic using the formula to get
cosx = (4√2 ± 0)/8 = √2/2
at this point you should realize that the equation would have factored to
(2cosx - √2)^2 = 0
then 2cosx = √2
and cosx = √2/2
so x = 45º or 315º (or pi/2 and 7pi/2 radians)
since it asked for all real values of x, we could give a general solution of
45º + 360kº or 135º + 360kº where k is an integer.
I will leave it up to you to give the general solution in radians
5+2\cos \left(2x\right)-4\sqrt{3}\cos \left(x\right)=0
To solve the equation 4 sin^2x + 4√2 cos x - 6 = 0, you can use the trigonometric identity sin^2x + cos^2x = 1 to rewrite it in terms of cos x.
Start by dividing the equation by 4 to simplify it:
sin^2x + √2 cos x - 3/2 = 0
Next, use the identity sin^2x = 1 - cos^2x to substitute sin^2x in the equation:
(1 - cos^2x) + √2 cos x - 3/2 = 0
Rearrange the terms:
- cos^2x + √2 cos x - 1/2 = 0
Now multiply each term by -1 to make it easier to solve:
cos^2x - √2 cos x + 1/2 = 0
To solve this quadratic equation, we can use the quadratic formula:
cos x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = -√2, and c = 1/2.
Substitute these values into the quadratic formula:
cos x = (-(-√2) ± √((-√2)^2 - 4(1)(1/2))) / (2 * 1)
cos x = (√2 ± √(2 - 2)) / 2
cos x = (√2 ± √0) / 2
The square root of 0 is 0, so simplify the expression:
cos x = (√2 ± 0) / 2
There are two solutions for cos x:
1. cos x = (√2 + 0) / 2 = (√2) / 2 ≈ 0.707
2. cos x = (√2 - 0) / 2 = (√2) / 2 ≈ 0.707
To find the corresponding values of x, use the inverse cosine function:
1. cos x = (√2) / 2
x = cos^(-1)((√2) / 2)
x = π/4 ≈ 0.785
2. cos x = (√2) / 2
x = cos^(-1)((√2) / 2)
x = -π/4 ≈ -0.785
So the solutions for the equation 4 sin^2x + 4√2 cos x - 6 = 0 for all real values of x are x = π/4 and x = -π/4.
To solve the equation 4sin^2x + 4√2cosx - 6 = 0, we can start by using the trigonometric identity sin^2x + cos^2x = 1.
First, let's replace sin^2x with (1 - cos^2x) in the equation:
4(1 - cos^2x) + 4√2cosx - 6 = 0
Expanding and simplifying the equation, we get:
4 - 4cos^2x + 4√2cosx - 6 = 0
-4cos^2x + 4√2cosx - 2 = 0
Next, let's divide the equation by 2 to simplify:
-2cos^2x + 2√2cosx - 1 = 0
Now, let's multiply the equation by -1 to change the signs:
2cos^2x - 2√2cosx + 1 = 0
We can now consider this equation as a quadratic equation in terms of cosx, where a = 2, b = -2√2, and c = 1. Applying the quadratic formula, we get:
cosx = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values of a, b, and c into the formula, we have:
cosx = (-(-2√2) ± √((-2√2)^2 - 4(2)(1))) / (2(2))
cosx = (2√2 ± √(8 - 16)) / 4
cosx = (2√2 ± √(-8)) / 4
Since we are solving for real values of x, we can see that √(-8) is not a real number. Therefore, there are no real solutions to this equation.
Hence, the original equation 4sin^2x + 4√2cosx - 6 = 0 has no real solutions for x.