Solve-sinx-cosx-1-x-R-

Tinku Tara June 4, 2023 Trigonometry 0 Comments

Facebook Tweet Pin

Question Number 14189 by tawa tawa last updated on 29/May/17

Solve: (√(sinx)) + (√(cosx)) = 1 x ∈ R

Solve : \sqrt{sinx} + \sqrt{cosx} = 1

x \in R

Answered by ajfour last updated on 29/May/17

let (√(sin x))=u and (√(cos x))=v this means u+v=1 ...(i) further as, sin^2 x+cos^2 x=1 u^4 +v^4 =1 ...(ii) we should know (u+v)^4 =u^4 +v^4 +uv(u^2 +v^2 ) +3uv(u+v)^2 using (i) and (ii): 1=1+uv(1−2uv)+3uv ⇒ uv(1−2uv+3)=0 uv=0 , or uv=2 since (√(sin x))(√(cos x))≠2, so uv=(√(sin x))(√(cos x))=0 ⇒ either sin x=0, or cos x=0 and since here sin x≥0 , cos x≥0 x=2nπ or x=2nπ+(π/2) .

l e t \sqrt{\sin x} = u a n d \sqrt{\cos x} = v

t h i s m e a n s u + v = 1 \dots (i)

f u r t h e r a s, \sin^{2} x + \cos^{2} x = 1

u^{4} + v^{4} = 1 \dots (i i)

w e s h o u l d k n o w

{(u + v)}^{4} = u^{4} + v^{4} + u v (u^{2} + v^{2})

+ 3 u v {(u + v)}^{2}

u s i n g (i) a n d (i i) :

1 = 1 + u v (1 - 2 u v) + 3 u v

\Rightarrow u v (1 - 2 u v + 3) = 0

u v = 0, o r u v = 2

s i n c e \sqrt{\sin x} \sqrt{\cos x} \neq 2, s o

u v = \sqrt{\sin x} \sqrt{\cos x} = 0

\Rightarrow e i t h e r \sin x = 0, o r \cos x = 0

a n d s i n c e h e r e \sin x ⩾ 0, \cos x ⩾ 0

x = 2 n π

o r x = 2 n π + \frac{π}{2} .

Commented by tawa tawa last updated on 29/May/17

Wow, God bless you sir.

Wow, God bless you sir .

Commented by mrW1 last updated on 29/May/17

x=2kπ or 2kπ+(π/2)

x = 2 k π o r 2 k π + \frac{π}{2}

Commented by tawa tawa last updated on 29/May/17

Sir. there appears to be a contradition from line 5 to 4 from the bottom.

Sir . there appears to be a contradition from line 5 to 4 from the bottom .

Commented by tawa tawa last updated on 29/May/17

Sir. can you explain ???.or it is correct !

Sir . can you explain ? ? ? . or it is correct!

Commented by mrW1 last updated on 29/May/17

if x=((nπ)/2) and n=3 cos x=−1 (√(cos x))=(√(−1)) !

i f x = \frac{n π}{2} a n d n = 3

\cos x = - 1

\sqrt{\cos x} = \sqrt{- 1}!

Commented by ajfour last updated on 29/May/17

very sorry .. i was elsewhere in my thoughts..

v e r y s o r r y . . i w a s e l s e w h e r e i n

m y t h o u g h t s . .

Commented by tawa tawa last updated on 29/May/17

God bless you sir.

God bless you sir .

Commented by tawa tawa last updated on 29/May/17

God bless you sir

God bless you sir

Commented by ajfour last updated on 29/May/17

Miss Tawa Q. 14194 upload again please..

M i s s T a w a Q . 14194 u p l o a d

a g a i n p l e a s e . .

Commented by tawa tawa last updated on 29/May/17

Have done that sir.

Have done that sir .

Commented by tawa tawa last updated on 29/May/17

Where have you been sir MRW1

Where have you been sir MRW 1

Commented by mrW1 last updated on 30/May/17

Dear Tawa: I′m still here. In recent time I watched more than give answers.

D e a r T a w a : I^{'} m s t i l l h e r e . I n r e c e n t

t i m e I w a t c h e d m o r e t h a n g i v e a n s w e r s .

Commented by ajfour last updated on 09/Jun/18

Glad to know that, Sir.

G l a d t o k n o w t h a t, S i r .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/May/17

sinx+cosx+2(√(sinx.cosx))=1 2(√(sinx.cosx))=1−(sinx+cosx) 4sinx.cosx=1+(sin^2 x+cos^2 x+2sinx.cosx)−2(sinx+cosx) ⇒sinx+cosx=1−sinx.cosx 1+2sinx.cosx=1−2sinx.cosx+sin^2 x.cos^2 x ⇒sinx.cosx(sinx.cosx−4)=0 ⇒ { ((sinx.cosx=4⇒sin2x=8 (impossible))),((sinx.cosx=0⇒sin2x=0⇒2x=2kπ)) :} ⇒x=kπ .■

s i n x + c o s x + 2 \sqrt{s i n x . c o s x} = 1

2 \sqrt{s i n x . c o s x} = 1 - (s i n x + c o s x)

4 s i n x . c o s x = 1 + ({s i n}^{2} x + {c o s}^{2} x + 2 s i n x . c o s x) - 2 (s i n x + c o s x)

\Rightarrow s i n x + c o s x = 1 - s i n x . c o s x

1 + 2 s i n x . c o s x = 1 - 2 s i n x . c o s x + {s i n}^{2} x . {c o s}^{2} x

\Rightarrow s i n x . c o s x (s i n x . c o s x - 4) = 0

\Rightarrow {\begin{cases} s i n x . c o s x = 4 \Rightarrow s i n 2 x = 8 (i m p o s s i b l e) \\ s i n x . c o s x = 0 \Rightarrow s i n 2 x = 0 \Rightarrow 2 x = 2 k π \end{cases}

\Rightarrow x = k π .

Commented by tawa tawa last updated on 29/May/17

God bless you sir.

God bless you sir .

Commented by mrW1 last updated on 29/May/17

sin 2x=0 but cos x≥0 ⇒x=2kπ and 2kπ+(π/2)

\sin 2 x = 0

b u t \cos x ⩾ 0

\Rightarrow x = 2 k π a n d 2 k π + \frac{π}{2}

Commented by tawa tawa last updated on 29/May/17

Where did you go sir MrW1, have not been seeing you online.

Where did you go sir MrW 1, have not been seeing you online .

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Facebook Tweet Pin

Leave a Reply Cancel reply