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If-6tan-2-x-tan-x-sec-x-sec-2-x-where-pi-2-lt-x-lt-3pi-2-has-the-roots-x-1-and-x-2-Find-the-value-of-sin-x-1-x-2-




Question Number 128645 by john_santu last updated on 09/Jan/21
 If 6tan^2 x+tan x.sec x = sec^2 x where  (π/2)<x<((3π)/2) has the roots x_1  and x_2 . Find  the value of sin (x_1 +x_2 ).
$$\:\mathrm{If}\:\mathrm{6tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{tan}\:\mathrm{x}.\mathrm{sec}\:\mathrm{x}\:=\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{where} \\ $$$$\frac{\pi}{\mathrm{2}}<\mathrm{x}<\frac{\mathrm{3}\pi}{\mathrm{2}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} .\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right). \\ $$
Answered by liberty last updated on 09/Jan/21
 (•) ((6sin^2 x)/(cos^2 x))+((sin x)/(cos^2 x))=(1/(cos^2 x))   (•) 6sin^2 x+sin x−1=0   (•)(3sin x−1)(2sin x+1)=0   (∵)  { ((sin x_1 =(1/3) ∧ sin x_2 =−(1/2) or)),((sin x_1 =−(1/2) ∧ sin x_2 =(1/3))) :}
$$\:\left(\bullet\right)\:\frac{\mathrm{6sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\left(\bullet\right)\:\mathrm{6sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\left(\bullet\right)\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2sin}\:\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\left(\because\right)\:\begin{cases}{\mathrm{sin}\:\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\:\wedge\:\mathrm{sin}\:\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}}\\{\mathrm{sin}\:\mathrm{x}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:\mathrm{sin}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$

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