The set of solutions of the equation ({3}-1) θ+({3}+1) θ=2 is

Correct Answer is : {2nπ±π4+π12:n∈Z}\{2nπ ± π/4 + π/12 : n {Z}\}{2nπ±4π+12π:n∈Z}

\{2nπ ± π/4 + π/12 : n {Z}\}

\{2nπ ± π/4 - π/12 : n {Z}\}

\{nπ + (-1)ⁿ π/4 + π/12 : n {Z}\}

\{nπ + (-1)ⁿ π/4 - π/12 : n {Z}\}

Correct Answer is : {2nπ±π4+π12:n∈Z}\{2nπ ± π/4 + π/12 : n {Z}\}{2nπ±4π+12π:n∈Z}

Test Index

EAMCET Engineering 2010 Solved Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2

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,

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

\sin \alpha = \frac{\sqrt{3}-1}{2\sqrt{2}}

\text{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)

\text{So, } \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 = 2n\pi \pm \frac{\pi}{4}, \theta = 2n\pi \pm \frac{\pi}{4}+\alpha

. . . (ii)

\because \cos 15 = \frac{\sqrt{3}+1}{2\sqrt{2}}

\text{ie, } \cos \alpha = \cos 15^{\circ} = \cos \frac{\pi}{12}

\Rightarrow \alpha = \frac{\pi}{12}

From Eq. (ii),

\left[\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}\right], n \in \mathbb{Z}

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