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Question: How do you find the general solution for \({{\tan }^{2}}x=3\)?...

How do you find the general solution for tan⁡2x=3{{\tan }^{2}}x=3?

Explanation

Solution

Now we know that trigonometric functions have repetitive values. Hence we get the same values for different inputs. Now we know that if x is the solution to any trigonometric equation then any coterminal angle of x is also the solution of the given equation. Hence to find the general solution we will write all the angles in general form. Now we know we can get coterminal angles by adding 2nπ2n\pi for different integers n. Hence we can find the general solution to any trigonometric equation by just finding one solution.

Complete step-by-step solution:
Now consider the given equation tan⁡2x=3{{\tan }^{2}}x=3.
Taking square root on both sides.
tan⁡x=±3\tan x=\pm \sqrt{3}
Hence we have tan⁡x=3\tan x=\sqrt{3} and tan⁡x=−3\tan x=-\sqrt{3}.
First let us consider tan⁡x=3\tan x=\sqrt{3}
Now we know that tan⁡(π3)=3\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3} .
Hence π3\dfrac{\pi }{3} is one of the solution of the equation tan⁡x=3\tan x=\sqrt{3}
Hence we can say that all the coterminal angles of π3\dfrac{\pi }{3} are also solution of the equation tan⁡x=3\tan x=\sqrt{3}
Hence the set \left\\{ \dfrac{\pi }{3}+2n\pi |n\in Z \right\\} is the solution to the equation tan⁡x=3\tan x=\sqrt{3} .
Now similarly consider tan⁡x=−3\tan x=-\sqrt{3}
We know that one of the solution of this equation is 2π3\dfrac{2\pi }{3}
Hence all the coterminal angles will also be the solution of the equation
Hence the set \left\\{ \dfrac{2\pi }{3}+2n\pi |n\in Z \right\\} is the solution to the equation tan⁡x=−3\tan x=-\sqrt{3}
Hence the solution to the equation tan⁡x=±3\tan x=\pm \sqrt{3} is \left\\{ \dfrac{\pi }{3}+2n\pi |n\in Z \right\\}\cup \left\\{ \dfrac{2\pi }{3}+2n\pi |n\in Z \right\\}
Hence the solution set of the equation tan⁡2x=3{{\tan }^{2}}x=3 is given by \left\\{ \dfrac{\pi }{3}+2n\pi |n\in Z \right\\}\cup \left\\{ \dfrac{2\pi }{3}+2n\pi |n\in Z \right\\}.

Note: Now note that while taking square root in any equation we need take both the cases negative and positive as x2=(−x)2{{x}^{2}}={{\left( -x \right)}^{2}} . also note that while writing coterminal angles we add 2nπ2n\pi where n is any integer and not natural number since we get coterminal angles by rotating the angle clockwise as well as anticlockwise. When we rotate clockwise we get x−2nπx-2n\pi and when we rotate anticlockwise we get x+2nπx+2n\pi where x is the original angle.