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Question: How do you solve \({\tan ^2}x - 3 = 0\) and find all the solutions in the interval \(0 \leqslant x <...

How do you solve tan⁡2x−3=0{\tan ^2}x - 3 = 0 and find all the solutions in the interval 0⩽x<3600 \leqslant x < 360 ?

Explanation

Solution

Firstly, add the constant 33 on both sides of the equation. Now, square root on both sides. The square on tan⁡x\tan x and the square root get canceled and on the RHS we will be left with the square root of 33. Now find the values of tan⁡x\tan x which will be equal to the square root of 33 and in the given interval 0⩽x<3600 \leqslant x < 360.

Complete step-by-step answer:
The given trigonometric expression is, tan⁡2x−3=0{\tan ^2}x - 3 = 0
Now add the constant 33 on both sides of the equation.
⇒tan⁡2x−3+3=0+3\Rightarrow {\tan ^2}x - 3 + 3 = 0 + 3
On evaluating we get,
⇒tan⁡2x=3\Rightarrow {\tan ^2}x = 3
Now apply square root on both sides of the equation.
We do this to cancel out the square on tan⁡x\tan x
⇒tan⁡2x=3\Rightarrow \sqrt {{{\tan }^2}x} = \sqrt 3
⇒tan⁡x=±3\Rightarrow \tan x = \pm \sqrt 3
The values of tan⁡x=3\tan x = \sqrt 3 occurs only when x=60∘,240∘x = 60^\circ ,240^\circ in the given interval, 0⩽x<3600 \leqslant x < 360
And another case which is, tan⁡x=−3\tan x = - \sqrt 3 occurs only when x=120∘,300∘x = 120^\circ ,300^\circ in the given interval, 0⩽x<3600 \leqslant x < 360
∴\therefore The solution of the trigonometric expression, tan⁡2x−3=0{\tan ^2}x - 3 = 0 is when x=60∘,120∘,240∘,300∘x = 60^\circ, 120^\circ, 240^\circ, 300^\circ

Additional information: Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into cos⁡θ\cos \theta or sin⁡θ\sin \theta. Then combine them into single fractions. Now it’s most likely to use Trigonometric identities for the transformations if there are any. Know when and where to apply the Subtraction-Addition formula.

Note:
Always check when the trigonometric functions are given in degrees or radians. Whenever we are canceling the square root with a square the solution will always be in the form of ±\pm . Always check where both the trigonometric functions become negative or positive. Most of the problems can easily be solved by memorizing Quotient identities and Subtraction-Addition identities.