Find x where pi <= x <= 2pi? 4 cos^2 x - 2= sec^2 x - tan^2 x

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Find x where pi <= x <= 2pi? 4 cos^2 x - 2= sec^2 x - tan^2 x

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Answer 1 · 2018-05-23T01:10:21

$\frac{4 π}{3}; \frac{11 π}{6}$ $(4pi)/3; (11pi)/6$

Explanation:

$4 {cos}^{2} x - 2 = {sec}^{2} x - {tan}^{2} x$ $4cos^2 x - 2 = sec^2 x - tan^2 x$ (1)
Develop the right side:
$R S = \frac{1}{{cos}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x} = \frac{1 - {sin}^{2} x}{{cos}^{2} x} = \frac{{cos}^{2} x}{{cos}^{2} x} = 1$ $RS = 1/(cos^2 x) - sin^2 x/(cos^2 x) = (1 - sin^2 x)/(cos^2 x) = cos^2 x/(cos^2 x) = 1$
The equation (1) becomes:
$4 {cos}^{2} x - 2 = 1$ $4cos^2 x - 2 = 1$
${cos}^{2} x = \frac{3}{4}$ $cos^2 x = 3/4$
$cos x = \pm \frac{\sqrt{3}}{2}$ $cos x = +- sqrt3/2$
a. $cos x = \frac{\sqrt{3}}{2}$ $cos x = sqrt3/2$
Trig table and unit circle give 2 solutions of x -->
$x = \pm \frac{π}{6}$ $x = +- pi/6$
b. $cos x = - \frac{\sqrt{3}}{2}$ $cos x = - sqrt3/2$ -->
$x = \pm \frac{2 π}{3}$ $x = +- (2pi)/3$
Inside the interval $(π, 2 π)$ $(pi, 2pi)$ , the answers are:
$x = - \frac{π}{6}$ $x = - pi/6$ , or $x = \frac{11 π}{6}$ $x = (11pi)/6$ (co-terminal)
$x = - \frac{2 π}{3}$ $x = - (2pi)/3$ , or $x = π + \frac{π}{3} = \frac{4 π}{3}$ $x = pi + pi/3 = (4pi)/3$ (co-terminal)

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Find x where pi <= x <= 2pi? 4 cos^2 x - 2= sec^2 x - tan^2 x

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