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

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

Explanation

Solution

In this question, we have to find the value of x of an equation. The equation given to us consists of trigonometric functions. Therefore, we will apply the trigonometric formulas to get the solution for the answer. We begin this problem by dividing both sides of the equation by cos⁡2x{{\cos }^{2}}x and then apply the trigonometric formula sin⁡2xcos⁡2x=tan⁡2x\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\tan }^{2}}x on the left-hand side of the equation and make further simplification. Then, we will take the square root of both sides of the equation, and therefore, we get 2 equations. So, we will solve them separately to get the solution to the problem.

Complete step by step answer:
According to the problem, we have to find the value of x.
The equation given to us is sin⁡2x=3cos⁡2x{{\sin }^{2}}x=3{{\cos }^{2}}x ---------- (1)
Thus, we will apply the trigonometric formulas to get the solution.
We first divide cos⁡2x{{\cos }^{2}}xon both sides in the equation (1), we get
⇒sin⁡2xcos⁡2x=3cos⁡2xcos⁡2x\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{3{{\cos }^{2}}x}{{{\cos }^{2}}x}
Thus, we will now apply the trigonometric formula sin⁡2xcos⁡2x=tan⁡2x\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\tan }^{2}}x on the left-hand side of the above equation, we get
⇒tan⁡2x=3cos⁡2xcos⁡2x\Rightarrow {{\tan }^{2}}x=\dfrac{3{{\cos }^{2}}x}{{{\cos }^{2}}x}
On further solving, we get
⇒tan⁡2x=3\Rightarrow {{\tan }^{2}}x=3
Now, we will take the square root on both sides of the equation, we get
⇒tan⁡2x=3\Rightarrow \sqrt{{{\tan }^{2}}x}=\sqrt{3}
Thus, we get
⇒tan⁡x=±3\Rightarrow \tan x=\pm \sqrt{3}
Therefore, from the above equation, we get two separate equations, that is
tan⁡x=3\tan x=\sqrt{3} ------ (2) and
tan⁡x=−3\tan x=-\sqrt{3} ----------- (3)
Now, we will solve equation (2), which is
tan⁡x=3\tan x=\sqrt{3}
Thus, we take tan⁡−1{{\tan }^{-1}} on both sides of the above equation, we get
tan⁡−1(tan⁡x)=tan⁡−1(3){{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\sqrt{\left( 3 \right)}
As we know, the trigonometric formula tan⁡−1(tan⁡a)=a{{\tan }^{-1}}\left( \tan a \right)=a , therefore the above equation will become
x=tan⁡−1(3)x={{\tan }^{-1}}\sqrt{\left( 3 \right)}
x=π3+nπx=\dfrac{\pi }{3}+n\pi where n is some integer
Now, we will solve equation (3), which is
tan⁡x=−3\tan x=-\sqrt{3}
Thus, we take tan⁡−1{{\tan }^{-1}} on both sides of the above equation, we get
tan⁡−1(tan⁡x)=tan⁡−1(−(3)){{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( -\sqrt{\left( 3 \right)} \right)
As we know, the trigonometric formula tan⁡−1(tan⁡a)=a{{\tan }^{-1}}\left( \tan a \right)=a , therefore the above equation will become
x=tan⁡−1(−(3))x={{\tan }^{-1}}\left( -\sqrt{\left( 3 \right)} \right)
x=2π3+nπx=\dfrac{2\pi }{3}+n\pi where n is some integer
Therefore, for the trigonometric equation sin⁡2x=3cos⁡2x{{\sin }^{2}}x=3{{\cos }^{2}}x , its general solution is the value of x, that is either x=π3+nπx=\dfrac{\pi }{3}+n\pi or x=2π3+nπx=\dfrac{2\pi }{3}+n\pi , where n is some integer.

Note: While solving this problem, do mention all the trigonometric formulas in the steps wherever they are applicable. Keep in mind that the tan function is a periodic function that implies its value repeats after some interval, that is why we add nπn\pi to the values of x. It means that nπn\pi is added after every interval.