Question

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

Answer 1

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 $ \dfrac{{\sin x}}{{\cos x}} $ which can be written as $ \tan x $ too. So, after modifying the equation we will get $ {\tan ^2}x = 3 $ which will mean $ \tan x = \sqrt 3 $ . Now, we know that $ \tan x $ has the value of $ \sqrt 3 $ at $ \theta = \dfrac{\pi }{3} $ in the first quadrant and $ \theta = \dfrac{{2\pi }}{3} $ in the second quadrant, $ \theta = \dfrac{{4\pi }}{3} $ in the third quadrant and $ \theta = \dfrac{{5\pi }}{3} $ in the fourth quadrant. $ \dfrac{{5\pi }}{3} $

Complete step-by-step answer:
The given equation we have is $ {\sin ^2}x = 3{\cos ^2}x $
Bringing all the trigonometric terms on one side of the equation
 $ \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = 3................(1) $
Now, we know that,
 $ \dfrac{{\sin x}}{{\cos x}} = \tan x $
Therefore, $ {\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2} = {\tan ^2}x $
So, equation (1) becomes
\[ {\tan ^2}x = 3 \\
\Rightarrow \tan x = \pm \sqrt 3 \;
\]
Now, looking at the value of $ \tan x $ we have to find the values of special angles where $ \tan x = \pm \sqrt 3 $
In the first quadrant:-
 $ \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 = - \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:-
\[ \tan x = \sqrt 3 ................\left[ {tan{\text{ }}is{\text{ }}positive{\text{ }}in{\text{ }}third{\text{ }}quadrant} \right] \\
\Rightarrow x = {\tan ^{ - 1}}\sqrt 3 \\
x = \dfrac{{4\pi }}{3} \;
\]
In the fourth quadrant:-
\[ \tan x = - \sqrt 3 ..................\left[ {tan{\text{ }}is{\text{ }}negative{\text{ }}in{\text{ }}fourth{\text{ }}quadrant} \right] \\
\Rightarrow x = {\tan ^{ - 1}}( - \sqrt 3 ) \\
x = \dfrac{{5\pi }}{3} \;
\]
Therefore, the solution of $ {\sin ^2}x = 3{\cos ^2}x $ is $ x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3} $
So, the correct answer is “ $ x = \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 $ \pm \sqrt 3 $ as tan’s value, we considered all four quadrants. If it would have been only $ \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 $ - \sqrt 3 $ then we would have considered only the 2nd and 4th quadrant as tan has negative value in these two quadrants only.