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


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

$Q . F i n d t h e n u m b e r o f s o l u t i o n o f t h d$ $e q u a t i o n t a n x + s e c x = 2 c o s x l y i n g i n$ $t h e i n t e r v a l [0, 2 π] ? ?$
	Answered by MJS last updated on 18/Feb/20

	$\tan x + \sec x = 2 \cos x$ $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$ $1 + \sin x - {2 \cos}^{2} x = 0$ $1 + s - 2 c^{2} = 0$ $1 + s - 2 (1 - s^{2}) = 0$ $2 s^{2} + s - 1 = 0$ $s^{2} + \frac{1}{2} s - \frac{1}{2} = 0$ $s = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} but - 1 ⩽ s ⩽ 1 \Rightarrow s = - \frac{1 + \sqrt{3}}{2}$ $\sin x = - \frac{1 + \sqrt{3}}{2} \Rightarrow 2 solutions for 0 ⩽ x ⩽ 2 π$
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