Solve-for-real-numbers-1-1-tan-4-x-1-10-2-1-3-tan-2-x-

Question Number 171872 by Shrinava last updated on 21/Jun/22

Solve for real numbers :

\frac{1}{1 + \tan^{4} x} + \frac{1}{10} = \frac{2}{1 + 3 \tan^{2} x}

Answered by aleks041103 last updated on 21/Jun/22

t = {t a n}^{2} x

\frac{1}{1 + t^{2}} + \frac{1}{10} = \frac{2}{1 + 3 t}

(1 + 3 t) + \frac{(1 + 3 t) (1 + t^{2})}{10} = 2 (1 + t^{2})

10 + 30 t + 1 + 3 t + t^{2} + 3 t^{3} = 20 + 20 t^{2}

3 t^{3} - 19 t^{2} + 33 t - 9 = 0, t > 0

3 t^{3} - t^{2} - 18 t^{2} + 33 t - 9 = 0

t^{2} (3 t - 1) - 18 t^{2} + 6 t + 27 t - 9 = 0

t^{2} (3 t - 1) - 6 t (3 t - 1) + 9 (3 t - 1) = 0

(3 t - 1) (t^{2} - 6 t + 9) = 0

(3 t - 1) {(t - 3)}^{2} = 0

t = \frac{1}{3}, 3

t g x = \pm \frac{\sqrt{3}}{3}, \pm \sqrt{3}

x = \pm \frac{π}{6} + k π; \pm \frac{π}{3} + k π

Commented by Shrinava last updated on 22/Jun/22

Cool dear professor thank you