What are the solutions to 10(sec^2)x+5(tan^2)x-15=0 over the interval (0,2pi)?

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What are the solutions to 10(sec^2)x+5(tan^2)x-15=0 over the interval (0,2pi)?

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Answer 1 · 2018-05-11T04:41:16

$\frac{π}{6}; \frac{2 π}{3}; \frac{7 π}{6}; \frac{5 π}{3}$ $pi/6; (2pi)/3; (7pi)/6; (5pi)/3$

Explanation:

$10 {sec}^{2} x + 5 {tan}^{2} x - 15 = 0$ $10sec^2 x + 5tan^2 x - 15 = 0$
Replace ${sec}^{2} x$ $sec^2 x$ by $(1 + {tan}^{2} x)$ $(1 + tan^2 x)$ from trig identity.
$10 (1 + {tan}^{2} x) + 5 {tan}^{2} x - 15 = 0$ $10(1 + tan^2 x) + 5tan^2 x - 15 = 0$
$10 {tan}^{2} x + 5 {tan}^{2} x - 5 = 0$ $10tan^2 x + 5tan^2 x - 5 = 0$
$15 {tan}^{2} x = 5$ $15tan^2 x = 5$ --> ${tan}^{2} x = \frac{1}{3}$ $tan^2 x = 1/3$
$tan x = \pm \frac{\sqrt{3}}{3}$ $tan x = +- sqrt3/3$
Trig table and unit circle give 4 solutions for tan x -->
a. $tan x = \frac{\sqrt{3}}{3}$ $tan x = sqrt3/3$ -->
$x = \frac{π}{6}$ $x = pi/6$ and $x = \frac{π}{6} + π = \frac{7 π}{6}$ $x = pi/6 + pi = (7pi)/6$
b. $tan x = - \frac{\sqrt{3}}{3}$ $tan x = - sqrt3/3$ -->
$x = \frac{2 π}{3}$ $x = (2pi)/3$ and $x = π + \frac{2 π}{3} = \frac{5 π}{3}$ $x = pi + (2pi)/3 = (5pi)/3$

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What are the solutions to 10(sec^2)x+5(tan^2)x-15=0 over the interval (0,2pi)?

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