Get the answer to "Solve tan⁡x+sec⁡2x=1"

High School
Geometry
Trigonometry
Trigonometric equations and identities

EunoR

Answered question

2020-11-24

Solve

\tan x + \sec 2 x = 1

Answer & Explanation

hajavaF

Skilled2020-11-25Added 90 answers

When $x = 0 : \tan (0) + \sec (2 \cdot 0) = 1$ so x=0 is a solution, because $\tan (0) = 0 and \cos (0) and \sec (0) = 1 .$
There is another solution at x=67.5 degrees. See below: If $y = \sin (x) or \tan (x) . \cos (2 x) = 1 - 2 \sin^{2} (x) = 1 - 2 y^{2}, \sec (2 x) = \frac{1}{1 - 2 \sin^{2} (x)}, \cos (x) = \sqrt{1 - y^{2}} . \sin = o p \frac{p}{h} y p = \frac{y}{1} s o a d j = \sqrt{1 - y^{2}} . \tan (x) = o p \frac{p}{a} d j = \frac{y}{\sqrt{1 - y^{2}}}$ .
Therefore, $\tan (x) + \sec (2 x) = \frac{y}{\sqrt{1 - y^{2}}} + \frac{1}{1 - 2 y^{2}} = 1 .$
$\frac{y}{\sqrt{1 - y^{2}}} = 1 - \frac{1}{1 - 2 y^{2}} = \frac{1 - 2 y^{2} - 1}{1 - 2 y^{2}} = - 2 \frac{y^{2}}{1 - 2 y^{2}} .$
y=0 is a solution $(\sin (x) = 0 s o x = 0)$ and:
$\frac{1}{\sqrt{1 - y^{2}}} = - 2 \frac{y}{1 - 2 y^{2}},$
$\sqrt{1 - y^{2}} = \frac{2 y^{2} - 1}{2} y .$
Squaring: $1 - y^{2} = \frac{4 y^{4} - 4 y^{2} + 1}{4} y^{2}, 4 y^{2} - 4 y^{4} = 4 y^{4} - 4 y^{2} + 1, 8 y^{4} - 8 y^{2} + 1 = 0 .$
So $y^{2} = (8 \pm \frac{\sqrt{64 - 32}}{16} = \frac{1}{2} \pm \frac{\sqrt{2}}{4}$

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