How do you solve #4sinx = 3tanx#?

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Answer 1 · 2015-05-13T17:56:32

4sin x = 3tan x

Call tan x/2 = t. Use the trig identities in terms of tan x/2.

#sin x = (2t)/(1 + t^2) and tan x = (2t)/(1 - t^2)#

#8t(1 - t^2) = 6t(1 + t^2)#
#8t - 8t^3 = 6t + 6t^3#

#f(t) = 14t^3 - 2t = 2t(7t^2 - 1) = 2t(sqr7.t - -1)(sqr7.t + 1)#

f(x) = 4sin x - 3tan x = 0 when:
a. t = tan x/2 = 0 -> x/2 = pi -> x = 2pi

b. #t = tan x/2 = 1/sqr7 = 0.38 = tan 20.70 -> x = 41.40 deg

c. #t = tan x/2 = -1/sqr7 = -0.38 = tan (-20.70) -> x = -41.40 deg

Check:
#x = 2pi -> sin x = 0; tan x = 0 # Correct
#x = 41.40 -> sin x = 0.66 -> 4sin x = 2.64 # -> #3tan x = 2.64.# OK

#x = -41.40 -> sin x = -0.66 --> 4sin x = -2.64 # -># 3tan x = -2.64 #. OK