How do you solve #sin4x+2sin2x=0#?

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Answer 1 · 2016-12-22T14:53:38

#{0 + 2pin, pi/2 + 2pin, pi + 2pin, (3pi)/2 + 2pin}#

Expand using #sin(A + B) = sinAcosB + cosAsinB#:

#sin(2x + 2x) + 2sin2x= 0#

#sin2xcos2x + sin2xcos2x + 2sin2x = 0#

#2sin2xcos2x + 2sin2x = 0#

#2sin2x(cos2x + 1) = 0#

Case 1: #2sin2x= 0#

Use #sin2theta = 2sinthetacostheta#:

#2(2sinxcosx) = 0#

#4sinxcosx= 0#

Whenever #sinx = 0# and whenever #cosx = 0#, the entire equation will equal #0#. So, #x = 0 + 2pin#, #pi/2 + 2pin#, #pi + 2pin#, (3pi)/2 + 2pin#.

Case 2: #cos2x + 1 = 0#

Use the identity #cos2x = 2cos^2x - 1#.

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

#2cos^2x = 0#

#cos^2x= 0#

#cosx =0#

#x = pi/2 + 2pin# and #(3pi)/2 + 2pin#

Hopefully this helps!