How do you solve #3cosx - sin2x = 0# from 0 to 2pi?

Question

9194 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-07T02:21:20

#3cosx-sin2x=0#

#=>3cosx-2sinxcosx=0#

#=>cosx(3-2sinx)=0#

So # cosx=0 or sinx=3/2>1" not possible"#

When #cosx=0=cos(pi/2)=cos(2pi-pi/2)#

Hence #x=pi/2or(3pi)/2#

Answer 2 · 2018-05-07T02:24:55

Values inconsistent.

#3 cos x - sin 2x = 0#

#3 cos x = sin 2x#

#3 cancel(cos x) = 2 sin x cancel(cos x)#

#3 = 2 sin x#

#sin x = 3/2#

As sine can have a value only between -1 & +1, values given in the sum are inconsistent.