How do you solve $csc^2x-2=0$ and find all solutions in the interval $[0,2pi)$ ?

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How do you solve $csc^2x-2=0$ and find all solutions in the interval $[0,2pi)$ ?

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Answer 1 · 2017-04-04T01:14:08

$pi/4; (3pi)/4; (5pi)/4; (7pi)/4$

Explanation:

$1/(sin^2 x) = 2$
$sin^2 x = 1/2$
$sin x = +- 1/(sqrt2) = +- sqrt2/2$
Trig table and unit circle gives 4 solutions:
a. $sin x = sqrt2/2$ -->
$x = pi/4$ and $x = (3pi)/4$
b. $sin x = - sqrt2/2$
$x = - pi/4$ , or $x = (7pi)/4$ (co-terminal), and
$x = pi - (-pi/4) = pi + pi/4 = (5pi)/4$

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How do you solve $csc^2x-2=0$ and find all solutions in the interval $[0,2pi)$ ?

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How do you solve csc^2x-2=0csc^2x-2=0 and find all solutions in the interval [0,2pi)[0,2pi)?

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How do you solve $csc^2x-2=0$ and find all solutions in the interval $[0,2pi)$ ?