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How do you solve #2sinx=sqrt3#?

Answer 1

Josiah Anderson

#(2pi)/3+2kpi, pi/3+2kpi# where #kinZZ#

#3sin(x)=sqrt3 rarr sin(x)=sqrt3/2#

#sin(x)=sqrt3/2# at #(2pi)/3+2kpi, pi/3+2kpi# where #kinZZ#

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Answer 2

Nolan Alexander

To solve (2\sin(x) = \sqrt{3}), divide both sides by 2, then take the inverse sine (or arcsin) of both sides. This yields (x = \frac{\pi}{3} + 2\pi n) and (x = \frac{2\pi}{3} + 2\pi n), where (n) is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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