How do you find all values of x in the interval [0, 2pi] in the equation #cos x =sin 2x#?

Question

5335 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T07:44:32

#S={pi/6, pi/2, (5pi)/6, (3pi)/2}#

#cosx - sin 2x =0#

#cosx - 2sin x cos x=0#

#cosx(1-2sinx)=0#

#cos x=0 or 1-2sinx =0#

#x=cos^-1 0 or -2sinx=-1-> sinx= 1/2-> x=sin^-1(1/2)#

#x=pi/2 , (3pi)/2 or x=pi/6 , (5pi)/6#

#S={pi/6, pi/2, (5pi)/6, (3pi)/2}#