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:

& \sin {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \\\ & \cos {{15}^{\circ }}=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\\ \end{aligned}$$ Then we will use the trigonometric identity to solve it further, which is given by, $$\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{aligned} & \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{aligned}$$ So we will divide the given equation by $$2\sqrt{2}$$. $$\begin{aligned} & \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{aligned}$$ As we know that $$\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{aligned} & \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{aligned}$$ $$\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.