The total number of solutions of the equation tan x + sec x = 2 which lie in the interval [0, 2π] is


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Question Number 17095 by Tinkutara last updated on 30/Jun/17

$The total number of solutions of the$ $equation \tan x + \sec x = 2 which lie in$ $the interval [0, 2 π] is$
	Answered by sma3l2996 last updated on 30/Jun/17
	$tanx+secx=2⇔tanx+(√(1+tan^2 x))=2 (2−tanx)^2 =1+tan^2 x 4+tan^2 x−4tanx=1+tan^2 x 4tanx=3⇔x=tan^(−1) ((3/4))+nπ \n∈N x=tan^(−1) ((3/4))+nπ \n=(0,1,2)$
	$t a n x + s e c x = 2 \Leftrightarrow t a n x + \sqrt{1 + {t a n}^{2} x} = 2$ ${(2 - t a n x)}^{2} = 1 + {t a n}^{2} x$ $4 + {t a n}^{2} x - 4 t a n x = 1 + {t a n}^{2} x$ $4 t a n x = 3 \Leftrightarrow x = {t a n}^{- 1} (\frac{3}{4}) + n π ∖ n \in N$ $x = {t a n}^{- 1} (\frac{3}{4}) + n π ∖ n = (0, 1, 2)$
	Commented by Tinkutara last updated on 01/Jul/17

	$Thanks Sir!$
	Commented by ajfour last updated on 01/Jul/17

	$but \tan x is - ve there . .$ $so \sec x + \tan x = \frac{5}{4} - \frac{3}{4} \neq 2 .$ $while squaring you gathered a false$ $root x = π - \sin^{- 1} \frac{3}{5}, reject that .$
	Commented by sma3l2996 last updated on 01/Jul/17

	$I t h i n k y o u d i d m i s t a k o n t h e l i n e 2$ $t h e c o r r e c t i s : s i n x + 1 = 2 c o s x$
	Commented by ajfour last updated on 01/Jul/17

	$why ?, you have θ = \sin^{- 1} (\frac{3}{5}) as a$ $true answer . so one solution .$
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