Can u please solve sin3x=4sin²x for 0<x<360?

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Answer 1 · 2017-07-20T16:38:24

#x=0^@# or #30^@# or #150^@# or#180^@#

As #sin3x=3sinx-4sin^3x#, the given equation becomes

#4sin^3x+4sin^2x-3sinx=0#

or #sinx(4sin^2x+4sinx-3)=0#

or #sinx(2sinx-1)(2sinx+3)=0#

As #2sinx+3>0#

either #sinx=0# i.e. #x=0^@# or #180^@#

or #sinx=1/2# i.e. #x=30^@# or #150^@#