Solveeit Logo
Login

Question

Question: Find the value of x if \(\sqrt{3}\sin x=\cos x\) ....

Find the value of x if 3sin⁡x=cos⁡x\sqrt{3}\sin x=\cos x .

Explanation

Solution

Hint : The Trigonometric ratios table helps to find the values of trigonometric standard angles such as 0∘,30∘,45∘,60∘ and 90∘0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ \text{ and 90}{}^\circ . So, use the trigonometric table to convert the equation given in the question to the form tanx=tan⁡αtanx=\tan \alpha and then use the formula of the general solution of tanx to get the answer.

Complete step by step solution :
Before moving to the solution, let us discuss the nature of sine and cosine function, which we would be using in the solution. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, and using the relations between the different trigonometric ratios, we get

Now to start with the solution to the above question, we will try to simplify the expression given in the question.
3sin⁡x=cos⁡x\sqrt{3}\sin x=\cos x
⇒3×sin⁡xcos⁡x=1\Rightarrow \sqrt{3}\times \dfrac{\sin x}{\cos x}=1
⇒sin⁡xcos⁡x=13\Rightarrow \dfrac{\sin x}{\cos x}=\dfrac{1}{\sqrt{3}}
We know tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} . So, our equation becomes:
tan⁡x=13\tan x=\dfrac{1}{\sqrt{3}}
Now from the trigonometric table, we know that tan⁡π6=13\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}} . So, our equation becomes:
tan⁡x=tan⁡π6\tan x=\tan \dfrac{\pi }{6}
We know that the general solution of the trigonometric equation tanx=tany⁡tanx=\operatorname{tany} is x=nπ+yx=n\pi +y .
Therefore, the general solution to our equation is x=nπ+π6x=n\pi +\dfrac{\pi }{6} , where n is an integer.

Note : Be careful about the calculation and the signs of the formulas you use as the signs in the formulas are very confusing and are very important for solving the problems. Also, it would help if you remember the properties related to complementary angles and trigonometric ratios. The above equation has infinite values of x satisfying the equation tan⁡x=tan⁡π6\tan x=\tan \dfrac{\pi }{6} , which is clear from the general solution. However, the principal values of x are π6 and 7π6\dfrac{\pi }{6}\text{ and }\dfrac{7\pi }{6} .