The general solution of the equation (√{3}-1) θ+(√{3}+1) θ=2 is

Correct Answer is : 2nπ±π/4+π/122nπ±π/4+π/{12}2nπ±π/4+π/12

Correct Answer is : 2nπ±π/4+π/122nπ±π/4+π/{12}2nπ±π/4+π/12

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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 23 of 160

Marks: +1, -0

(√{3}-1) \sin θ+(√{3}+1) \cos θ=2

Solution:

Given,

(√{3}-1) \sin θ+(√{3}+1) \cos θ=2

Let

\;\; √{3}-1=r \sin α

. . . (i)

and

√{3}+1=r \cos α

∴ r^{2}(\sin ^{2} α+\cos ^{2} α)=(√{3}-1)^{2}+(√{3}+1)^{2}

⇒ r^{2}=3+1-2 √{3}+3+1+2 √{3}

⇒ \;\; r^{2}=8 ⇒ r=2 √{2}

and

\tan α=\;{√{3}-1}/{√{3}+1}=\;{1-1 \/ √{3}}/{1+1 \/ √{3}}

\;\; =\tan (\;{π}/{4}-\;{π}/{6})=\tan (\;{π}/{12}) ⇒ α=\;{π}/{12}

From Eq. (i), we have

r \sin α \sin θ+r \cos α \cos θ=2

⇒ r \cos (θ-α) =2

⇒ 2 √{2} \cos (θ-α) =2

⇒ \cos (θ-α) =\;{1}/{√{2}}

⇒ \;\; θ-α =2 n π ± \;{π}/{4}

⇒ \;\; θ=2 n π ± \;{π}/{4}+\;{π}/{12}

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