Question

Find the general solution of the equation $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$.
A. $2n\pi \pm \dfrac{\pi }{4}+\dfrac{\pi }{12}$
B. $n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}+\dfrac{\pi }{12}$
C. $2n\pi \pm \dfrac{\pi }{4}-\dfrac{\pi }{12}$
D. $n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{12}$

Answer 1

Hint: In this question, we need to use the angle sum property of a triangle according to which the sum of all the angles of a triangle is $\pi $. A trigonometric equation typically has several solutions or an infinite number of solutions because all trigonometric ratios are periodic in nature. So, applying trigonometric tan techniques, we will simplify the given equation to obtain the final equation.

Formula Used:The trigonometric formulas of tangent are:
$\begin{align}
  & \tan (\pi -\theta )=-\tan \theta \\
 & \tan (B+C)=\dfrac{\tan B+\tan C}{1-\tan B\tan C} \\
 & \tan {{90}^{\circ }}=\infty \\
\end{align}$

Complete step- by- step solution:If the equation involves a variable $0\le x<2\pi $, then the solutions are called principal solutions. An equation involving trigonometric functions of unknown angles, as $\cos B=\dfrac{1}{2}$. The principal solution of the equation $\sin x=\dfrac{1}{2}$ that is less than $\dfrac{\pi }{2}$.
The general solution of $\sin x=-1$ is$n\pi +{{(-1)}^{n}}2\pi $
Let $\sqrt{3}+1=r\cos \alpha $ and $\sqrt{3}-1=r\sin \alpha $
Then $r=\sqrt{{{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}}=2\sqrt{2}$
$\begin{align}
  & \tan \alpha =\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}} \\
 & \Rightarrow \tan \alpha =\dfrac{1-\tan \dfrac{\pi }{6}}{1+\tan \dfrac{\pi }{6}} \\
 & \Rightarrow \tan \alpha =\tan \left( \dfrac{\pi }{4}-\dfrac{\pi }{6} \right) \\
 & \Rightarrow \alpha =\dfrac{\pi }{12} \\
\end{align}$
The given equation is reduced to
$\begin{align}
  & 2\sqrt{2}\cos (\theta -\alpha )=2 \\
 & \Rightarrow \cos \left( \theta -\dfrac{\pi }{12} \right)=\cos \dfrac{\pi }{4} \\
 & \Rightarrow \theta -\dfrac{\pi }{12}=2n\pi \pm \dfrac{\pi }{4} \\
 & \Rightarrow \theta =2n\pi \pm \dfrac{\pi }{4}+\dfrac{\pi }{12} \\
\end{align}$
We know that,
$\cos (A-B)=\cos A\cos B+\sin A\sin B$ and $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
Hence the correct option is A.

Option ‘A’ is correct

Note: The generic solution often contains arbitrary constants (in the case of an ODE) or arbitrary functions, and it encompasses all feasible solutions (in the case of a PDE.) By dividing the angle into a sum of a bigger angle and the acute angle, the principal value of the trigonometric function can be determined. After determining the acute angle value for the trigonometric function, we can go on to determine the trigonometric function's primary solution.