Question

How do you find the solution of $\sqrt 3 \sin x - \cos x = 1$?

Answer 1

Hint: Since this question is in geometric form, first try to find out if the given equation is similar to any of the trigonometric identities. If it is similar try to reconstruct the given equation into the suitable identity which will enable us to further solve the question and reach our desired solution.

Formula used:
The identity used in this sum is $\sin (a - b) = \sin (a)\cos (b) - \cos (a)\sin (b)$.

Complete step by step solution:
In this kind of situation always try to make some sort of adjustment in your question such that it turns into some of our trigonometric identities.
If you notice carefully in this question, we can use the identity $\sin (a - b) = \sin (a)\cos (b) - \cos (a)\sin (b)$ but we will have to make some adjustments.
First we will divide our entire equation by 2
$\dfrac{{\sqrt 3 \sin x - \cos x}}{2} = \dfrac{1}{2}$
Solving further we get
$\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}$
To make use of the above identity we need to find an angle of cos which has it’s value$\dfrac{{\sqrt 3 }}{2}$ and an angle of sin which has it’s value as $\dfrac{1}{2}$.
Also we need to make sure that both these angles are the same.
Clearly we know that $\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$ and $\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$.
Hence we have got the values of $\dfrac{{\sqrt 3 }}{2}$ and $\dfrac{1}{2}$.
Now substituting these values in equation (1) we get
$\cos \left( {\dfrac{\pi }{6}} \right)\sin (x) - \sin \left( {\dfrac{\pi }{6}} \right)\cos (x) = \dfrac{1}{2}$
Rearranging the above equation we get,
$\sin (x)\cos \left( {\dfrac{\pi }{6}} \right) - \cos (x)\sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$
Hence we have the left hand side of the above equation in the form of the identity we were supposed to use. Hence we have
$\sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}$
Taking the sine inverse function on both sides of the equation we get
$\left( {x - \dfrac{\pi }{6}} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
We know that the value of ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ is $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$
therefore value of $\left( {x - \dfrac{\pi }{6}} \right)$ will be $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$
Hence, $\left( {x - \dfrac{\pi }{6}} \right) = \left\{\begin{matrix}
  \dfrac{\pi }{6} + 2n\pi \\
  \dfrac{{5\pi }}{6} + 2n\pi \\
\end{matrix} \right.$
$ \Rightarrow x = \left\{\begin{matrix}
   \dfrac{\pi }{3} + 2n\pi & \\
   \pi + 2n\pi &
\end{matrix}\right.$ is the general solution.

Note: In this kind of sums we try to bring the trigonometric equations into basic trigonometric identities as it makes it easier to solve the trigonometric identities. If it is not possible to bring the entire question in basic identity if we at least bring some part of the equation in some trigonometric equation it becomes helpful.