Find the sum of the three smallest positive values of x such that 4cos^2(2x-pi) = 3. (Answer in radians.)
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
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
so we have the first four values of
x = π/12, 5π/12 , 7π/12, and 11π/12
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
ahhh, missed the part where we want the sum of the first 3
π/12 + 5π/12 + 7π/12 = 13π/12
To find the sum of the three smallest positive values of x that satisfy the equation 4cos^2(2x-π) = 3, we need to solve the equation for x and find the values that satisfy the condition. Here's how you can do it:
Step 1: Rewrite the equation: 4cos^2(2x-π) - 3 = 0.
Step 2: Use the double angle identity for cosine: cos(2θ) = 2cos^2(θ) - 1.
Step 3: Apply the double angle identity to the equation: 4(2cos^2(2x-π) - 1)^2 - 3 = 0.
Step 4: Simplify the equation: 4(4cos^4(2x-π) - 4cos^2(2x-π) + 1) - 3 = 0.
Step 5: Expand and rearrange: 16cos^4(2x-π) - 16cos^2(2x-π) + 1 = 0.
Step 6: Let's substitute y = cos(2x-π) to simplify the equation. Now we have: 16y^4 - 16y^2 + 1 = 0.
Step 7: This is now a quadratic equation in terms of y^2. Solve the quadratic equation: 16y^2 - 16y^2 + 1 = 0.
Step 8: Factor the quadratic equation: (4y^2 - 1)^2 = 0.
Step 9: Solve for y: 4y^2 - 1 = 0.
Step 10: Solve for y: y^2 = 1/4.
Step 11: Take the square root of both sides: y = ±(1/2).
Step 12: Substitute back y = cos(2x-π) and solve for x:
For y = 1/2:
cos(2x-π) = 1/2.
Solving for 2x-π:
2x-π = ±π/3 + 2kπ (where k is an integer).
Solving for x:
x = π/6 + kπ/2 (where k is an integer).
For y = -1/2:
cos(2x-π) = -1/2.
Solving for 2x-π:
2x-π = ±2π/3 + 2kπ (where k is an integer).
Solving for x:
x = π/3 + kπ/2 (where k is an integer).
Step 13: Find the three smallest positive values of x:
For k = 0: x = π/6, π/3.
For k = 1: x = π/2.
Step 14: Calculate the sum of the three smallest positive values of x:
Sum = π/6 + π/3 + π/2
Step 15: Simplify the sum:
Sum = π/6 + 2π/6 + 3π/6
Step 16: Add the fractions:
Sum = 6π/6
Step 17: Simplify the fraction:
Sum = π
Therefore, the sum of the three smallest positive values of x that satisfy the equation 4cos^2(2x-π) = 3 is π radians.