Question #37b3d

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Question #37b3d

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Answer 1 · 2016-08-18T04:43:53

$\pm \frac{π}{3} + 2 k π$ $+- pi/3 + 2kpi$

Explanation:

Simplify both sides by ${cos}^{2} t$ $cos^2 t$ . (Condition cos x different to 0)
The remainder equation is:
$\frac{1}{{cos}^{2} t} + 5 = 3 \frac{{sin}^{2} t}{{cos}^{2} t}$ $1/(cos^2 t) + 5 = 3(sin^2 t)/(cos^2 t)$
$\frac{1 + 5 {cos}^{2} t}{{cos}^{2} t} = \frac{3 {sin}^{2} t}{{cos}^{2} t}$ $(1 + 5cos^2 t)/(cos^2 t) = (3sin^2 t)/(cos^2 t)$
Simplify both sides by ${cos}^{2} t .$ $cos^2 t.$
Replace in the equation $3 {sin}^{2} t$ $3sin^2 t$ by $(3 - 3 {cos}^{2} t)$ $(3 - 3cos^2 t)$ -->
$1 + 5 {cos}^{2} t = 3 - 3 {cos}^{2} t$ $1 + 5cos^2 t = 3 - 3cos^2 t$
$8 {cos}^{2} t = 2$ $8cos^2 t = 2$ --> ${cos}^{2} t = \frac{2}{8} = \frac{1}{4}$ $cos^2 t = 2/8 = 1/4$
$cos t = \pm \frac{1}{2}$ $cos t = +- 1/2$
Trig table and unit circle -->
$cos t = \pm \frac{1}{2}$ $cos t = +- 1/2$ --> $t = \pm \frac{π}{3}$ $t = +- pi/3$
General answer:
$t = \pm \frac{π}{3} + 2 k π$ $t = +- pi/3 + 2kpi$

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Question #37b3d

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