how do you solve for x?
2cos^24x-1=0
First find the cos of 24
multiply this by 2
add 1 to the other side
Finally, divide that answer with 1
and... u have the answer!!! whatever that is...
I am reading your question as
2cos^2 (4x) - 1 = 0
so cos^2 (4x) = 1/2
cos 4x = ±1/√2
so 4x must be an angle in each of the 4 quadrants.
but I know cos 45º = 1/√2
so 4x = 45 or 4x = 135 or 4x = 225 or 4x = 315
then
x = 11.25 or x = 33.75 or x=56.25 or x=78.75
the period of cos 4x = 90º
so by adding/subtracting multiples of 90 to any of the above answers will produce other answers
e.g. 33.75 + 90 = 123.75 would be another possible answer
proof:
LS
= 2 cos^2 (4*123.75) - 1
= 2cos^2 (495) - 1
= 2(1/2) - 1
= 0 = RS
just follow my way, its much simpler
MC, except you don't get the right answer, lol
prove it Reiny
:)
my way always works
To solve for x in the equation 2cos^2(4x) - 1 = 0, you can follow these steps:
Step 1: Add 1 to both sides of the equation:
2cos^2(4x) = 1
Step 2: Divide both sides of the equation by 2:
cos^2(4x) = 1/2
Step 3: Take the square root of both sides of the equation:
cos(4x) = ±√(1/2)
Step 4: Find the angle whose cosine value is ±√(1/2):
cos(45°) = √(2)/2
cos(225°) = -√(2)/2
Step 5: Set up the equations with the found angles:
4x = 45° + 360°n, where n is an integer
4x = 225° + 360°n, where n is an integer
Step 6: Solve for x:
x = (45° + 360°n) / 4, where n is an integer
x = (225° + 360°n) / 4, where n is an integer
These equations give you the solutions for x in the given equation. Keep in mind that there are infinitely many solutions due to the periodic nature of the cosine function. You can substitute different integer values for n to find additional values of x.