Solve-the-following-inequality-sinx-1-cosx-1-where-0-x-lt-2pi-cosx-0-

Question Number 1407 by 112358 last updated on 29/Jul/15

S o l v e t h e f o l l o w i n g i n e q u a l i t y

\frac{s i n x + 1}{c o s x} ⩽ 1

w h e r e 0 ⩽ x < 2 π, c o s x \neq 0

Commented by 123456 last updated on 29/Jul/15

f (x) = \frac{\sin x + 1}{\cos x}

f (x) ⩽ 1 \Leftrightarrow f (x) - 1 ⩽ 0

g (x) = f (x) - 1 = \frac{\sin x + 1}{\cos x} - 1 = \tan x + \sec x - 1

g (x) ⩽ 0

Commented by 112358 last updated on 30/Jul/15

x \in (\frac{π}{2}, 2 π) \cap x \neq \frac{3 π}{2} i s c o r r e c t .

Commented by 123456 last updated on 29/Jul/15

x \in (π / 2, 2 π) ∖ {3 π / 2}

Commented by 112358 last updated on 30/Jul/15

G o o \overset{}{d} d e d u c t i o n!

Commented by Rasheed Ahmad last updated on 31/Jul/15

S o r r y I a c c i d e n t l y d e l e t e d m y

c o m m e n t! T h e r e i s n o w a y t o

r e c o v e r! I w i l l r e w r i t e i t s o o n .

Answered by Rasheed Ahmad last updated on 03/Aug/15

(R a s h e e d S o o m r o)

C a s e - I

L e t x i s i n Q - I

c o s x > 0, s i n x > 0

\frac{s i n x + 1}{c o s x} ⩽ 1 \Rightarrow s i n x + 1 ⩽ c o s x

\Rightarrow s i n x ⩽ c o s x - 1

\Rightarrow 1 ⩽ - (\frac{1 - c o s x}{s i n x})

\Rightarrow 1 ⩽ - t a n \frac{x}{2} \dots . [∵ \frac{1 - c o s x}{s i n x} = t a n \frac{x}{2}]

- 1 ⩾ t a n \frac{x}{2}

\Rightarrow t a n \frac{x}{2} ⩽ - 1 (i)

B u t t a n \frac{x}{2} > 0 [\frac{x}{2} i s i n Q - I] (i i)

(i) a n d (i i) a r e c o n t r a d i c t o r y .

T h e y {c a n}^{'} t b e t r u e s i m u l t a n e o u s l y .

H e n c e i n Q - I t h e r e i s n o s o l u t i o n .

x \notin (0, \frac{π}{2})

★ ★ ★

C a s e - I I

L e t x i s i n Q - I I

s i n x > 0, c o s x < 0

s i n x + 1 ⩾ c o s x

s i n x ⩾ c o s x - 1

1 ⩾ - (\frac{1 - c o s x}{s i n x})

1 ⩾ - t a n \frac{x}{2} \Rightarrow - 1 ⩽ t a n \frac{x}{2}

t a n \frac{x}{2} ⩾ - 1 \dots \dots \dots \dots \dots \dots (i)

N o w (x i s i n Q - I I) \Rightarrow (\frac{x}{2} i s i n Q - I)

S o, t a n \frac{x}{2} > 0 \dots \dots \dots \dots \dots (i i)

(i) a n d (i i) h a v e t a n \frac{x}{2} > 0 c o m m o n f o r

w h i c h b o t h a r e t r u e a t s a m e t i m e .

S o x \in (\frac{π}{2}, π)

★ ★ ★

C a s e - I I I

L e t x i s i n Q - I I I

s i n x < 0, c o s x < 0

S i m i l a r w a y o f d e d u c t i o n l e a d s u s :

t a n \frac{x}{2} ⩽ - 1 \dots \dots (i)

N o w

(x i s i n Q - I I I) \Rightarrow (\frac{x}{2} i s i n Q - I I)

S o t a n \frac{x}{2} < 0 \dots \dots . (i i)

H e r e i n t e r s e c t i o n o f (i) a n d (i i)

i s t a n \frac{x}{2} ⩽ - 1

S o t h e s o l u t i o n o f g i v e n i n e q u a l i t y

i s^{'} s o l u t i o n o f t a n \frac{x}{2} ⩽ - 1^{'}

t a n \frac{x}{2} ⩽ - 1 \Rightarrow \frac{x}{2} ⩽ \frac{3 π}{4}

\Rightarrow x ⩽ \frac{3 π}{2}

B u t c o s \frac{3 π}{2} = 0

S o f i n a l y x < \frac{3 π}{2}

T h i s c o v e r s w h o l e Q - I I I

x \in (π, \frac{3 π}{2})

★ ★ ★

C a s e - I V

L e t x i s i n Q - I V

s i n x < 0, c o s x > 0

I n s i m i l a r w a y a s a b o v e w e d e d u c e :

t a n \frac{x}{2} ⩾ - 1 \dots \dots \dots \dots \dots \dots (i)

N o w

(x i s i n Q - I V) \Rightarrow (\frac{x}{2} i s i n Q - I I)

H e n c e t a n \frac{x}{2} < 0 \dots \dots \dots . . (i i)

(i) a n d (i i) m a y b e w r i t t e n a s :

- 1 ⩽ t a n \frac{x}{2} < 0

t a n \frac{x}{2} \in [- 1, 0)

\frac{x}{2} \in [\frac{3 π}{4}, π)

x \in [\frac{3 π}{2}, 2 π)

T h a t i s w h o l e Q - I V i n c l u d i n g

\frac{3 π}{2} b u t e x c u d i n g 2 π .

B u t c o s \frac{3 π}{2} = 0 s o i t a l s o b e e x c l u d e d .

H e n c e x \in (\frac{3 π}{2}, 2 π)

★ ★ ★

C a s e - V

L e t x i s a q u a r d a n t a l a n g l e

F o r x = \frac{π}{2} o r \frac{3 π}{2} c o s x = 0 w h i c h i s

a g a i n s t t h e r e s t r i c t i o n c o s x \neq 0 .

F o r x = 0 o r 2 π t h e g i v e n i n e q u a l i t y

h o l d s b u t 2 π i s o u t o f d o m a i n .

H e n c e o n l y 0 o u t o f q u a r d a n t a l

a n g l e s i s i n c l u d e d i n s o l u t i o n .

Thus,

Final Answer i s :

x \in {0} \cup (\frac{π}{2}, 2 π) - {\frac{3 π}{2}}

Commented by 123456 last updated on 03/Aug/15

f (x) = \frac{\sin x + 1}{\cos x}

\lim_{x \to 3 π / 2} f (x) = 0 < 1

this is a curious fact :)

Commented by Rasheed Ahmad last updated on 03/Aug/15

A n d t h a t m e a n s f (x) i s

d i s c o n t i n u o u s a t x = \frac{3 π}{2} .