Question

Solve the equation \[\left( \sqrt{3}-1 \right)\cos \theta +\left( \sqrt{3}+1 \right)sin\theta =2\].

Answer 1

Hint: In the above question, we will first of all we will divide the given equation by \[2\sqrt{2}\] and then we will substitute the values of \[\sin {{15}^{\circ }}\] and \[\cos {{15}^{\circ }}\] which is as follows:
\[\begin{align}
  & \sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \\
 & \cos {{15}^{\circ }}=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\
\end{align}\]
Then we will use the trigonometric identity to solve it further, which is given by,
\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\].

Complete Step-by-step answer:
We have been given the equation \[\left( \sqrt{3}-1 \right)\cos \theta +\left( \sqrt{3}+1 \right)sin\theta =2\].
To solve the given equation in the form of \[a\cos \theta +b\sin \theta =c\], we divide the equation by \[\sqrt{{{a}^{2}}+{{b}^{2}}}\].
\[\begin{align}
  & \Rightarrow \sqrt{{{\left( \sqrt{3}-1 \right)}^{2}}+{{\left( \sqrt{3}+1 \right)}^{2}}}=\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\sqrt{3}+{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}+2\sqrt{3}} \\
 & =\sqrt{3+1+3+1-2\sqrt{3}+2\sqrt{3}}=\sqrt{8}=2\sqrt{2} \\
\end{align}\]
So we will divide the given equation by \[2\sqrt{2}\].
\[\begin{align}
  & \dfrac{\left( \sqrt{3}-1 \right)}{2\sqrt{2}}\cos \theta +\dfrac{\left( \sqrt{3}+1 \right)}{2\sqrt{2}}sin\theta =\dfrac{2}{2\sqrt{2}} \\
 & \dfrac{\left( \sqrt{3}-1 \right)}{2\sqrt{2}}\cos \theta +\dfrac{\left( \sqrt{3}+1 \right)}{2\sqrt{2}}sin\theta =\dfrac{1}{\sqrt{2}} \\
\end{align}\]
As we know that \[\cos {{15}^{\circ }}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}\] and \[\sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].
So by substituting these values in the above equation, we get as follows:
\[\sin {{15}^{\circ }}\cos \theta +\cos {{15}^{\circ }}sin\theta =\dfrac{1}{\sqrt{2}}\].
We know that \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\].
We also know that \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\].
So by using this trigonometric identity to the above equation, we get as follows:
\[\sin \left( {{15}^{\circ }}+\theta \right)=\sin {{45}^{\circ }}\].
We already know that the general solution for \[\sin \theta =\sin \alpha \] is \[\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha \], where ‘n’ is any integer, therefore we get,
\[\begin{align}
  & \sin \left( {{15}^{\circ }}+\theta \right)=\sin {{45}^{\circ }} \\
 & {{15}^{\circ }}+\theta =n\pi +{{\left( -1 \right)}^{n}}{{45}^{\circ }} \\
 & \dfrac{\pi }{12}+\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4} \\
\end{align}\]
\[\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{12}\], where n is any integer.
Therefore, the solution of the given equation is \[\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{12}\].

Note: Be careful while doing calculation, specially while dividing the equation by \[2\sqrt{2}\] and substituting the values of \[\cos {{15}^{\circ }}\] and \[\sin {{15}^{\circ }}\]. Also, you can calculate the value of \[\cos {{15}^{\circ }}\] and \[\sin {{15}^{\circ }}\] by using the trigonometric identity \[\cos 2\theta =2co{{s}^{2}}\theta -1\] here by substituting \[\theta ={{15}^{\circ }}\] we get the value of \[\cos {{15}^{\circ }}\]. Also, check the final solution as once by substituting the value of ‘\[\theta \]’ in the given equation.