How would you solve for x in this equation? 2tanx = cos2x

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Answer 1 · 2017-08-08T10:19:03

See below.

We have #cos(2x)=1-2sin^2x# then

#2sinx/cosx=1-2sin^2x# or

#4sin^2x = (1-2sin^2x)^2(1-sin^2x)# now calling #y = sin^2x# we have

#4y=(1-2y)^2(1-y)# or

#9 y - 8 y^2 + 4 y^3-1=0# with a real root

#y = 1/6 (4 - 11/root(3)(6 sqrt[177]-71)+ root(3)(6 sqrt[177]-71)) approx 0.123914#

then #y=sin^2x=0.123914# and finally

#x = arcsin(pmsqrt(1/6 (4 - 11/root(3)(6 sqrt[177]-71)+ root(3)(6 sqrt[177]-71)) ))+2kpi#