Solve each equaton for all values of x cos6x cos3x 0Solutio

Solve each equaton for all values of x: cos(6x)- cos(3x) = 0

Solution

cos(6x) - cos(3x) = 0

we have cos(2x) = 2cos²x - 1

==> cos(2 (3x)) - cos(3x) = 0

==> 2cos²(3x) - 1 - cos(3x) = 0

==> 2cos²(3x) - cos(3x) -1 = 0

==> 2cos²(3x) - 2cos(3x) +cos(3x) -1 = 0

==> 2cos(3x) (cos(3x) - 1) + 1(cos(3x) -1) = 0

==> (cos(3x) - 1)(2cos(3x)+ 1) = 0

==> cos(3x) - 1 = 0 or 2cos(3x)+ 1 = 0

==> cos(3x) = 1 , cos(3x) = -1/2

==> 3x = 2n (or) (2n +1) - (/3) (or) (2n - 1) + (/3)

==> x = 2n/3 (or)   (1/3) [(2n +1) - (/3)] (or)   (1/3) [(2n - 1) + (/3)]

==> x = 2n/3 (or)   (1/3) [ 2n + - /3] (or)   (1/3) [2n - + /3]

==> x = 2n/3 (or)   (1/9) [ 6n + 2] (or)   (1/9) [6n - 2]

==> x = 2n/3 (or) 2(3n + 1)/9 (or) 2(3n - 1)/9

Hence solutions are x = 2n/3 , x = 2(3n + 1)/9 , x = 2(3n - 1)/9 , n belongs to integers