How do you find all values of x in the interval [0, 2pi] in the equation #tan 2x= sqrt(2) - cos 2x#?

1 Answer
A. S. Adikesavan
Aug 1, 2018

#x = kpi +- 1/2 arccos (0.8832), kpi +- 1/2 arccos (- 0.4690 )#,
#k = 0, +-1, +-2, +-3, ...#

Explanation:

If #c = cos 2x, tan 2x = +- 1/c sqrt ( 1- c^2 #.

The given equation reads

#+- 1/c sqrt ( 1- c^2 ) = sqrt 2 - c# giving

#c^4 - 2 sqrt2 c^3 + 3 c^2 - 1 = 0#

Graph locates c near 1 and -0.5.
graph{y-(x^4-sqrt8 x^3+3 x^2-1)=0}
Trial and error graphical method, with astute scaling gives

4-sd# c = cos 2x = 0.8832 and -0 4690#
graph{y-(x^4-sqrt8 x^3+3 x^2-1)=0[0.883 0.88341 -0.001 0.001]}
graph{y-(x^4-sqrt8 x^3+3 x^2-1)=0[-0.469 -0.46895 -0.001 0.001]}

Use #x = 1/2 ( 2kpi +- arccos c), k = 0, +- 1, +- 2, +- 3, ...#

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