The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is
The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is
- $\left\{2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}: n \in Z\right\}$
- $\left\{2 n \pi \pm \frac{\pi}{4}-\frac{\pi}{12}: n \in Z\right\}$
- $\left\{n \pi+(-1)^n \frac{\pi}{4}+\frac{\pi}{12}: n \in Z\right\}$
- $\left\{n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{12}: n \in Z\right\}$
Solution
$(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$
$\frac{\sqrt{3}-1}{2} \sin \theta+\frac{\sqrt{3}+1}{2} \cos \theta=1$ ...(i)
Comparing with $a \sin \theta+b \cos \theta=1$.
ie, $\quad a=\frac{\sqrt{3}-1}{2}, b=\frac{\sqrt{3}+1}{2}$
$\sqrt{a^2+b^2}=\sqrt{\frac{(\sqrt{3}-1)^2}{4}+\frac{(\sqrt{3}+1)^2}{4}}$
$=\frac{1}{2} \sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}}$
$=\frac{1}{2} \sqrt{8}=\frac{1}{2} \cdot 2 \sqrt{2}=\sqrt{2}$
Dividing on both sides by $\sqrt{2}$ in Eq. (i), we get
$\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \sin \theta+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \cos \theta=\frac{1}{\sqrt{2}}$
Let $\quad \sin \alpha=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Then, $\cos \alpha=\sqrt{1-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2}$
$=\sqrt{1-\frac{(4-2 \sqrt{3})}{8}}$
$=\sqrt{\frac{8-4+2 \sqrt{3}}{8}}=\sqrt{\frac{4+2 \sqrt{3}}{8}}$
$=\sqrt{\frac{\sqrt{3}+1}{4}}=\sqrt{\frac{(\sqrt{3}+1)^2}{8}}$
$=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)$
So, $\quad \sin \alpha \cdot \sin \theta+\cos \alpha \cdot \cos \theta=\frac{1}{\sqrt{2}}$
$\cos (\theta-\alpha)=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}$
$\theta-\alpha=2 n \pi \pm \frac{\pi}{4}, \theta=2 n \pi \pm \frac{\pi}{4}+\alpha$ $\ldots$ (ii)
$\because \quad \cos 15=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
ie, $\quad \cos \alpha=\cos 15^{\circ}=\cos \frac{\pi}{12}$
$\Rightarrow \quad \alpha=\frac{\pi}{12}$
From Eq. (ii), $\left[\theta=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}\right] . n \in Z$
Asked in: AP EAMCET 2010
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