How do you solve $2cosx + sqrt3=0$ for x in the interval [0,2pi)?

Question

How do you solve $2cosx + sqrt3=0$ for x in the interval [0,2pi)?

1 Answer

Impact of this question

26803 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-15T05:32:01

$(5pi)/6 and (7pi)/6$

Explanation:

$2cos x + sqrt3 = 0$
$cos x = -sqrt3/2$
Unit circle and trig table of special arcs give
$cos x = -sqrt3/2$ --> $x = +- (5pi)/6$
$-(5pi)/6$ --> co-terminal with $(7pi)/6$
Answers: $(5pi)/6 and (7pi)/6$ (0 < x < $2pi$ )

Science

Math

Humanities

... and beyond

How do you solve $2cosx + sqrt3=0$ for x in the interval [0,2pi)?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2cosx + sqrt3=02cosx + sqrt3=0 for x in the interval [0,2pi)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2cosx + sqrt3=0$ for x in the interval [0,2pi)?