Solve the equation #5sin4x=2sin2x# if #0^@ <= x <= 360^@#?

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Answer 1 · 2016-10-17T10:47:01

Solution is #{0^o,39.232^o,90^o,140.768^o,180^o,219.232^o,270^o,320.768^o,360^o}#

#5sin4x=2sin2x#

#hArr5xx2sin2xcos2x=2sin2x#

or #10sin2xcos2x-2sin2x=0#

or #2sin2x(5cos2x-1)=0#

Hence either #sin2x=0# i.e. #2x={0^o,180^o,360^o,540^o,720^o,....}#

or in the given range #0<=x<=360^o#, #x={0^o,90^o,180^o,270^o,360^o}#

or #5cos2x-1=0# i.e. #cos2x=1/5=cos(+-78.463^o)# i.e.

#2x={78.463^o,(360^o-78.463^o),(360^o +78.463^o),(720^o-78.463^o)}#

#x={39.232^o,140.768^o,219.232^o,320.768^o}#

Hence, solution is #{0^o,39.232^o,90^o,140.768^o,180^o,219.232^o,270^o,320.768^o,360^o}#
graph{5sin(4x)-2sin(2x) [-1, 7.26, -4, 6]}