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Question Number 82115 by Khyati last updated on 18/Feb/20

Q. Find the number of solution of thd equation tanx + secx = 2 cosx lying in the interval [0, 2π] ??

$${Q}.\:{Find}\:{the}\:{number}\:{of}\:{solution}\:{of}\:{thd} \\ $$$${equation}\:{tanx}\:+\:{secx}\:=\:\mathrm{2}\:{cosx}\:{lying}\:{in} \\ $$$${the}\:{interval}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:?? \\ $$

Answered by MJS last updated on 18/Feb/20

tan x +sec x =2cos x ((sin x)/(cos x))+(1/(cos x))=2cos x 1+sin x −2cos^2 x =0 1+s−2c^2 =0 1+s−2(1−s^2 )=0 2s^2 +s−1=0 s^2 +(1/2)s−(1/2)=0 s=−(1/2)±((√3)/2) but −1≤s≤1 ⇒ s=−((1+(√3))/2) sin x =−((1+(√3))/2) ⇒ 2 solutions for 0≤x≤2π

$$\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\:=\mathrm{2cos}\:{x} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\mathrm{2cos}\:{x} \\ $$$$\mathrm{1}+\mathrm{sin}\:{x}\:−\mathrm{2cos}^{\mathrm{2}} \:{x}\:=\mathrm{0} \\ $$$$\mathrm{1}+{s}−\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{1}+{s}−\mathrm{2}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{2}{s}^{\mathrm{2}} +{s}−\mathrm{1}=\mathrm{0} \\ $$$${s}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${s}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{but}\:−\mathrm{1}\leqslant{s}\leqslant\mathrm{1}\:\Rightarrow\:{s}=−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:{x}\:=−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi \\ $$

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