math (trig)
posted by Crystal .
how do you solve for x?
2cos^24x1=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
Respond to this Question
Similar Questions

math
how would you solve for x: (3)^.5sin(2x)2cos(2x)+2sin^2(x)=1 using trig identites. please help 
algebra
Did I do this problem the correct way if so is there a shorter way to explain this? 
Trig
Solve 2cos sq x = 1 using the unit circle. 
Trig
Solve the following trig equation 2cos(xπ/6)=1 
trig
solve 2cos^2xsinx=1, 0 < or equal to x < 2pie 
trig
Solve the equation on the interval [0,360) cos^2t+2cos(t)+1=0 
trig
Find minimum value of 2cos x + 2cos y 
trig
Find minimum value of 2cos x + 2cos y 
trig
Find minimum value of 2cos x + 2cos y 
alg2 check?
solve 3(8x2)=462(12x+1) 24x6=4624x+2 24x6=4824x 48x6=48 48x=42 x=42/48 or 7/8 what where the two mistakes made in the problem?