Question

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

Answer 1

`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, ...`