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

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