Solve the equation giving the exact solutions which lie in 0

Solution

1) cos(2x)= cos(x)

2cos^2x - 1 = cosx

2cos^2x - 1 - cosx = 0

2cos^2x -2cosx +cosx -1 =0

2cosx( cosx-1) +1(cosx -1) =0

(2cosx +1)(cosx -1) =0

cosx = -1/2

x = 2pi/.3 , 4pi/3

cosx =1

x = 0

Solution : x= 0, 2pi/3 , 4pi/3

2)3cos(2x)= sin(x)+2

3(1 -2sin^2x) = sinx +2

-6sin^2x -sinx +1 =0

-6sin^2x -3sinx +2cosx +1=0

-3sinx( 2sinx +1) +1(2cosx +1) =0

( 1-3sinx)(1+2cosx) =0

sinx = 1/3 ; x= 0.339 , pi -0.339 = 2.80

x = 0.339 , 2.80 radians

cosx = -1/2

x = 2pi/.3 , 4pi/3

Solution : x= 2pi/.3 , 4pi/3, 0.339 , 2.80