SOLUTION: sin (2x) = cos (2x) between (0, 2pi)

Question 551337: sin (2x) = cos (2x) between (0, 2pi)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
sin (2x) = cos (2x) between (0, 2pi)
-------
sin^2 = cos^2 = 1 - sin^2
2sin^2 = 1
sin^2(2x) = 1/2

2x = pi/4, 3pi/4, 5pi/4, 7pi/4, 9pi/4, 11pi/4, 13pi/4, 15pi/4
x = pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi,8
-----------
x = pi/8 + n*pi/4, n = 0 to 7

RELATED QUESTIONS
sin^2x-cos^2x=cosx between intervals... (answered by josgarithmetic)

Find all the solutions between 0 and 2pi to the equation... (answered by ewatrrr)

Solve on the interval [0, 2pi] Sin*2x - sin x + 1=... (answered by lwsshak3)

Solve sin 2x = sin x for x between 0 and 2pi. (answered by stanbon)

find all solutions between 0 2pi for 7... (answered by lwsshak3)

Find all solutions of each of the equations in the interval (0, 2pi) cos(x- pi/2) +... (answered by lwsshak3)

Solve for x between 0 and 2pi (both included): sin(4x) = cos(2x). I changed sin(4x)... (answered by Alan3354)

Solve sin 2x + cos 2x =... (answered by Alan3354,ikleyn)

Solve sin (2x) + cos (2x) =... (answered by Alan3354,ikleyn)