The-total-number-of-solutions-of-the-equation-tan-x-sec-x-2-which-lie-in-the-interval-0-2pi-is-

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

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 .

Facebook Tweet Pin