In0,π2, one solution of 3+1cosecx+3−1secx=42 is 5π12. If other solution is λπ36 then λ=

3+1cosecx+3−1secx=42⇒3+1cosx+3−1sinx=42sinxcosx⇒3+122cosx+3−122sinx=2sinxcosx⇒sinx+5π12=sin2x⇒2x=nπ+−1nx+5π12⇒x=5π12 or x=7π36⇒λ=7

In0,π2, one solution of 3+1cosecx+3−1secx=42 is 5π12. If other solution is λπ36 then λ=

3+1cosecx+3−1secx=42⇒3+1cosx+3−1sinx=42sinxcosx⇒3+122cosx+3−122sinx=2sinxcosx⇒sinx+5π12=sin2x⇒2x=nπ+−1nx+5π12⇒x=5π12 or x=7π36⇒λ=7

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In $(0, \frac{π}{2})$ , one solution of $(\sqrt{3} + 1) \cos e c x + (\sqrt{3} - 1) \sec x =$ $4 \sqrt{2}$ is $\frac{5 π}{12}$ . If other solution is $\frac{λ π}{36}$ then $λ$ =

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Detailed Solution

$\begin{array}{l} (\sqrt{3} + 1) \cos e c x + (\sqrt{3} - 1) \sec x = 4 \sqrt{2} \\ \Rightarrow (\sqrt{3} + 1) \cos x + (\sqrt{3} - 1) \sin x = 4 \sqrt{2} \sin x \cos x \\ \Rightarrow \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cos x + \frac{\sqrt{3} - 1}{2 \sqrt{2}} \sin x = 2 \sin x \cos x \\ \Rightarrow \sin (x + \frac{5 π}{12}) = \sin 2 x \\ \Rightarrow 2 x = n π + {(- 1)}^{n} (x + \frac{5 π}{12}) \\ \Rightarrow x = \frac{5 π}{12} o r x = \frac{7 π}{36} \\ \Rightarrow λ = 7 \end{array}$