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hichkok12 [17]
3 years ago
9

Precalculus only right answer please and thank you. Help me

Mathematics
1 answer:
Deffense [45]3 years ago
5 0

Given:

\frac{2tanx}{1-tan^2(x)} =\sqrt{3}

We know the identity

tan(2x)= \frac{2tanx}{1-tan^2(x)}

So we can equate tan(2x) =\sqrt{3}

tan(x) =\sqrt{3} when x=\frac{\pi}{3}

Tan is positive in first and third quadrant

So we will get one move value for x

tan(x) =\sqrt{3} when x=\frac{4\pi}{3}

So for  tan(2x) =\sqrt{3}

2x=\frac{4\pi}{3}  and 2x=\frac{\pi}{3}

Divide by 2 on both sides

x=\frac{2\pi}{3}  and 2x=\frac{\pi}{6}

To get general solution we add n\pi

So option A  and option C are correct.

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