Question

Find the sum of the three smallest positive values of x such that 4cos^2(2x-pi) = 3. (Answer in radians.)

Answer 1

cos^2 (2x-pi) = .75
cos (2x-pi) = +/-.866

let z = 2x-pi
if z = 30 degrees (pi/3) yes
if z = 150 degrees (2pi/3) yes
if z = -30 degrees(-pi/3) yes
if z = -150 degrees(-2pi/3)

if z = -pi/3
then 2 x-pi =-pi/3
2 x = 2 pi/3
x = pi/3 , well that is +

if z = +pi/3
2 x = pi + pi/3 = 4 pi/3
x = 2 pi/3

if z = -2 pi/3
2 x = pi - 2 pi/3
2 x = pi/3
x = pi/6
so maybe pi/6 , pi/3 , 2 pi/3

Answer 2

4cos^2(2x-pi) = 3
cos^2 (2x - pi) = 3/4

cos (2x-pi) = ± √3/2
I know that cos 30° = √3/2
so 2x - pi = pi/6
2x = pi + pi/6
x = 7pi/12 ---->105°

since we had cos(2x-pi) = ± √3/2
2x - pi could have been in any of the 4 quadrants

2x - pi = pi - pi/6 ---> in II
2x = 11pi/6
x = 11pi/12 ----> 165 °

2x-pi = pi+pi/6
2x = 13pi/6
x = 13pi/12 -----> 195 °

2x-pi = 2pi - pi/6
2x = 17pi/6
x = 17pi/12 ----> 255 °

notice that the difference between the first and 3rd is 90 ° or pi/2
and the difference between the 4th and the 2nd is 90 °
so we can back up 90 ° from 165 ° and get another positive smaller value of x of
75 ° which would be 5pi/12
and 105 - 90 = 15
so the first one is π/12

<b>so we have the first four values of
x = π/12, 5π/12 , 7π/12, and 11π/12</b>

check Wolfram:
https://www.wolframalpha.com/input/?i=4cos%5E2(2x-pi)+%3D+3

showing the first 3 of those.
If you hover your cursor over the red intersection point, it will show
(.2618,3), (1.309,3), and (1.8326, 3)
which match my first 3 solutions for x
e.g
π/12 = .261799...
5π/12 =1.309
etc

Answer 3

ahhh, missed the part where we want the sum of the first 3

π/12 + 5π/12 + 7π/12 = 13π/12