How do you solve $sec (2 x + \frac{π}{3}) = \frac{2}{\sqrt{3}}$ $sec(2x + pi/3) = 2/sqrt(3)$ ?

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How do you solve $sec (2 x + \frac{π}{3}) = \frac{2}{\sqrt{3}}$ $sec(2x + pi/3) = 2/sqrt(3)$ ?

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Answer 1 · 2016-04-30T16:50:12

$- \frac{π}{12} and - \frac{π}{4}$ $-pi/12 and -pi/4$

Explanation:

Transform the equation to variable cos (2x + pi/3):
$\frac{1}{cos (\frac{2 x + π}{3})} = \frac{2}{\sqrt{3}}$ $1/(cos ((2x + pi)/3)) = 2/sqrt3$ . Cross mulyiply:
$2 cos (2 x + \frac{π}{3}) = \sqrt{3}$ $2cos (2x + pi/3) = sqrt3$
$cos (2 x + \frac{π}{3}) = \frac{\sqrt{3}}{2}$ $cos (2x + pi/3) = sqrt3/2$
Trig table and unit circle -->
$2 x + \frac{π}{3} = \pm \frac{π}{6}$ $2x + pi/3 = +- pi/6$
a. $2 x + \frac{π}{3} = \frac{π}{6} - \to 2 x = \frac{π}{6} - \frac{π}{3} = - \frac{π}{6} - \to x = - \frac{π}{12}$ $2x + pi/3 = pi/6 --> 2x = pi/6 - pi/3 = - pi/6 --> x = - pi/12$
b. $2 x + \frac{π}{3} = - \frac{π}{6} \to 2 x = - \frac{π}{6} - \frac{π}{3} = - \frac{π}{2} - \to x = - \frac{π}{4}$ $2x + pi/3 = -pi/6 -> 2x = - pi/6 - pi/3 = - pi/2 --> x = -pi/4$

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How do you solve $sec (2 x + \frac{π}{3}) = \frac{2}{\sqrt{3}}$ $sec(2x + pi/3) = 2/sqrt(3)$ ?

1 Answer

Explanation:

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How do you solve sec(2x+π3)=2√3 sec(2x + pi/3) = 2/sqrt(3)?

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How do you solve $sec (2 x + \frac{π}{3}) = \frac{2}{\sqrt{3}}$ $sec(2x + pi/3) = 2/sqrt(3)$ ?