Q-Find-the-number-of-solution-of-thd-equation-tanx-secx-2-cosx-lying-in-the-interval-0-2pi-

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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