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Question: How do you solve \( {\sin ^2}x = 3{\cos ^2}x \) ? \( \)...

How do you solve sin⁡2x=3cos⁡2x{\sin ^2}x = 3{\cos ^2}x ?

Explanation

Solution

Hint : Bring trigonometric terms on one side to simplify the equation and solve
First step we need to do to solve this question is to bring the trigonometric terms on one side of the equation. This step will give us the LHS in terms of sin⁡xcos⁡x\dfrac{{\sin x}}{{\cos x}} which can be written as tan⁡x\tan x too. So, after modifying the equation we will get tan⁡2x=3{\tan ^2}x = 3 which will mean tan⁡x=3\tan x = \sqrt 3 . Now, we know that tan⁡x\tan x has the value of 3\sqrt 3 at θ=π3\theta = \dfrac{\pi }{3} in the first quadrant and θ=2π3\theta = \dfrac{{2\pi }}{3} in the second quadrant, θ=4π3\theta = \dfrac{{4\pi }}{3} in the third quadrant and θ=5π3\theta = \dfrac{{5\pi }}{3} in the fourth quadrant. 5π3\dfrac{{5\pi }}{3}

Complete step-by-step answer :
The given equation we have is sin⁡2x=3cos⁡2x{\sin ^2}x = 3{\cos ^2}x
Bringing all the trigonometric terms on one side of the equation
sin⁡2xcos⁡2x=3................(1)\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = 3................(1)
Now, we know that,
sin⁡xcos⁡x=tan⁡x\dfrac{{\sin x}}{{\cos x}} = \tan x
Therefore, (sin⁡xcos⁡x)2=tan⁡2x{\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2} = {\tan ^2}x
So, equation (1) becomes

⇒tan⁡x=±3  \Rightarrow \tan x = \pm \sqrt 3 \;

Now, looking at the value of tan⁡x\tan x we have to find the values of special angles where tan⁡x=±3\tan x = \pm \sqrt 3
In the first quadrant:-
tan⁡x=+3..................[tan is positive in first quadrant] ⇒x=tan⁡−13 x=π3  \tan x = + \sqrt 3 ..................\left[ {tan{\text{ }}is{\text{ }}positive{\text{ }}in{\text{ }}first{\text{ }}quadrant} \right] \\\ \Rightarrow x = {\tan ^{ - 1}}\sqrt 3 \\\ x = \dfrac{\pi }{3} \;
In the second quadrant:-
tan⁡x=−3..................[tan is negative in second quadrant] ⇒x=tan⁡−1(−3) x=2π3  \tan x = - \sqrt 3 ..................\left[ {tan{\text{ }}is{\text{ }}negative{\text{ }}in{\text{ }}second{\text{ }}quadrant} \right] \\\ \Rightarrow x = {\tan ^{ - 1}}( - \sqrt 3 ) \\\ x = \dfrac{{2\pi }}{3} \;
In the third quadrant:-

⇒x=tan⁡−13 x=4π3  \Rightarrow x = {\tan ^{ - 1}}\sqrt 3 \\\ x = \dfrac{{4\pi }}{3} \;

In the fourth quadrant:-

⇒x=tan⁡−1(−3) x=5π3  \Rightarrow x = {\tan ^{ - 1}}( - \sqrt 3 ) \\\ x = \dfrac{{5\pi }}{3} \;

Therefore, the solution of sin⁡2x=3cos⁡2x{\sin ^2}x = 3{\cos ^2}x is x=π3,2π3,4π3,5π3x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}
So, the correct answer is “ x=π3,2π3,4π3,5π3x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3} ”.

Note : The negative and positive values of tan in the first, second, third and fourth quadrant should be memorized and used accordingly. Since the answer had ±3\pm \sqrt 3 as tan’s value, we considered all four quadrants. If it would have been only 3\sqrt 3 then we would have considered only the 1st and 3rd quadrant as tan has positive value in these two quadrants only. And suppose it would have had only −3- \sqrt 3 then we would have considered only the 2nd and 4th quadrant as tan has negative value in these two quadrants only.