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


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 14189 by tawa tawa last updated on 29/May/17

$Solve : \sqrt{sinx} + \sqrt{cosx} = 1$ $x \in R$
	Answered by ajfour last updated on 29/May/17

	$l e t \sqrt{\sin x} = u a n d \sqrt{\cos x} = v$ $t h i s m e a n s u + v = 1 . . . (i)$ $f u r t h e r a s, \sin^{2} x + \cos^{2} x = 1$ $u^{4} + v^{4} = 1 . . . (i i)$ $w e s h o u l d k n o w$ ${(u + v)}^{4} = u^{4} + v^{4} + u v (u^{2} + v^{2})$ $+ 3 u v {(u + v)}^{2}$ $u s i n g (i) a n d (i i) :$ $1 = 1 + u v (1 - 2 u v) + 3 u v$ $\Rightarrow u v (1 - 2 u v + 3) = 0$ $u v = 0, o r u v = 2$ $s i n c e \sqrt{\sin x} \sqrt{\cos x} \neq 2, s o$ $u v = \sqrt{\sin x} \sqrt{\cos x} = 0$ $\Rightarrow e i t h e r \sin x = 0, o r \cos x = 0$ $a n d s i n c e h e r e \sin x ⩾ 0, \cos x ⩾ 0$ $x = 2 n π$ $o r x = 2 n π + \frac{π}{2} .$
	Commented by tawa tawa last updated on 29/May/17

	$Wow, God bless you sir .$
	Commented by mrW1 last updated on 29/May/17

	$x = 2 k π o r 2 k π + \frac{π}{2}$
	Commented by tawa tawa last updated on 29/May/17

	$Sir . there appears to be a contradition from line 5 to 4 from the bottom .$
	Commented by tawa tawa last updated on 29/May/17

	$Sir . can you explain ? ? ? . or it is correct!$
	Commented by mrW1 last updated on 29/May/17

	$i f x = \frac{n π}{2} a n d n = 3$ $\cos x = - 1$ $\sqrt{\cos x} = \sqrt{- 1}!$
	Commented by ajfour last updated on 29/May/17

	$v e r y s o r r y . . i w a s e l s e w h e r e i n$ $m y t h o u g h t s . .$
	Commented by tawa tawa last updated on 29/May/17

	$God bless you sir .$
	Commented by tawa tawa last updated on 29/May/17

	$God bless you sir$
	Commented by ajfour last updated on 29/May/17

	$M i s s T a w a Q . 14194 u p l o a d$ $a g a i n p l e a s e . .$
	Commented by tawa tawa last updated on 29/May/17

	$Have done that sir .$
	Commented by tawa tawa last updated on 29/May/17

	$Where have you been sir MRW 1$
	Commented by mrW1 last updated on 30/May/17

	$D e a r T a w a : I^{'} m s t i l l h e r e . I n r e c e n t$ $t i m e I w a t c h e d m o r e t h a n g i v e a n s w e r s .$
	Commented by ajfour last updated on 09/Jun/18

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

	$God bless you sir .$
	Commented by mrW1 last updated on 29/May/17

	$\sin 2 x = 0$ $b u t \cos x ⩾ 0$ $\Rightarrow x = 2 k π a n d 2 k π + \frac{π}{2}$
	Commented by tawa tawa last updated on 29/May/17

	$Where did you go sir MrW 1, have not been seeing you online .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum