How do you solve #4tanx + sin2x = 0#?

2 Answers
Equivirial
May 4, 2015

Any integer multiple of #\pi#, [ ex) # x=0,\pi,-pi,2pi,-2pi, ... ]# will solve the equation.

Use that #tanx=sinx/cosx# and that #(sin2x=2sinx cosx)#

Making these substitutions,

#4tanx+sin2x=4sinx/cosx+2sinxcosx=#

#=sinx(4/cosx+2cosx)#

One way this expression can equal zero is if #sin x=0#. #sin x# will be zero whenever #x# is an integer multiple of #pi#.

Are there any other solutions?

The expression will also be zero when #4/cosx+2cosx=0#.

The preceding equation can be rearranged into the form,

#(cosx)^2=-2#

Because the square of a real number cannot be negative there are no solutions to this equation.

Therefore, integer multiples of #pi# are the only solutions.

Nghi N.
May 5, 2015

There is another way that gives same answer.
Call tan x = t . Use trig identity: # sin 2x = (2t)/(1 + t^2)#

#f(x) = 4t + (2t)/(1 + t^2) = [(2t).(3 + 2t^2)]/(1 + t^2)# = 0
The 2 quantities with t^2 are always positive.
f(x) = 0 when t = tan x = 0 -> x = 0 or #x = K.Pi#

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