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)#?

1 Answer

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