Question-14078

Question Number 14078 by Tinkutara last updated on 27/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17

4 {c o s}^{3} x - 3 c o s x - \sqrt{3} s i n x . c o s x + {c o s}^{2} x - \frac{1}{2} = - 2

8 {c o s}^{3} x + 2 {c o s}^{2} x - 2 \sqrt{3} s i n x . c o s x - 6 c o s x + 3 = 0

c o s x = t \Rightarrow

8 t^{3} + 2 t^{2} - 6 t + 3 = 2 \sqrt{3} t \sqrt{1 - t^{2}}

64 t^{6} + 4 t^{4} + 36 t^{2} + 9 + 32 t^{5} - 96 t^{4} + 48 t^{3} - 24 t^{3} + 12 t^{2} - 36 t = 12 t^{2} - 12 t^{4}

\Rightarrow 64 t^{6} + 32 t^{5} - 80 t^{4} + 24 t^{3} + 36 t^{2} - 36 t + 9 = 0

\Rightarrow t = c o s x = \frac{1}{2} \Rightarrow x = 2 k π \pm \frac{π}{3} .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17

Commented by ajfour last updated on 28/May/17

∙ \sin [2 (2 k π - \frac{π}{3}) - \frac{7 π}{6}]

= \sin (4 k π - \frac{11 π}{6}) = \sin \frac{π}{6} \neq - 1 .

∙ \sin [2 (2 k π + \frac{π}{3}) - \frac{7 π}{6}]

= \sin (4 k π - \frac{π}{2}) = - 1 .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/May/17

c o s (- x) = c o x (x) = \frac{1}{2}

Answered by ajfour last updated on 27/May/17

x = 2 n π + \frac{π}{3} .

Commented by Tinkutara last updated on 27/May/17

Can you explain the solution please ?

Answered by ajfour last updated on 27/May/17

a b o v e e q u a t i o n i m p l i e s t h a t

\cos 3 x = - 1 a n d \sin (2 x - \frac{7 π}{6}) = - 1

\Rightarrow 3 x = π + 2 k π; k \in Z

x = \frac{π}{3} + k (\frac{2 π}{3}) \dots . . (i)

a l s o 2 x - \frac{7 π}{6} = 2 n π - \frac{π}{2}

\Rightarrow x = \frac{π}{3} + n π \dots . . (i i)

f o r e q n s . (i) a n d (i i) t o b e

s i m u l t a n e o u s l y t r u e :

\boldsymbol x = 2 \boldsymbol n π + \boldsymbol π / 3 .

Commented by Tinkutara last updated on 27/May/17

But since \cos 3 x = - 1 = \cos π,

∴ 3 x = 2 n π \pm π

and similarly \sin x = \sin y implies

x = n π + {(- 1)}^{n} y

Why you {don}^{'} t use these conditions ?

Commented by ajfour last updated on 27/May/17

Commented by mrW1 last updated on 27/May/17

\cos (a) ⩾ - 1

\sin (b) ⩾ - 1

\Rightarrow \cos (a) + \sin (b) ⩾ - 2

“ =^{″} i s v a l i d o n l y i f

\cos (a) = - 1

\sin (b) = - 1

Commented by RasheedSindhi last updated on 27/May/17

a b o v e e q u a t i o n \overset{?}{i m p l i e s} t h a t

\cos 3 x = - 1 a n d \sin (2 x - \frac{7 π}{6}) = - 1

- - - - - - - - - - - - -

Why the above eq implies ?

a + b = c [= d + e (say)]

\Rightarrow a = d \land b = e ?

Answered by mrW1 last updated on 27/May/17

s i n c e - 2 ⩽ \cos (a) + \sin (b) ⩽ 2

f o r \cos 3 x + \sin (2 x - \frac{7 π}{6}) = - 2

\Rightarrow \cos 3 x = - 1

\Rightarrow \sin (2 x - \frac{7 π}{6}) = - 1

\Rightarrow 3 x = (2 n + 1) π, n \in Z

x = \frac{2 n + 1}{3} π = \frac{2 n π}{3} + \frac{π}{3}

\Rightarrow 2 x - \frac{7 π}{6} = 2 m π - \frac{π}{2}

\Rightarrow x = m π + \frac{π}{3}

\frac{2 n}{3} = m

\Rightarrow n = 3 k, k \in Z

S o l u t i o n i s

x = 2 k π + \frac{π}{3}

Commented by Tinkutara last updated on 28/May/17

But if \cos x = \cos y, we should use

x = 2 n π \pm y and for \sin x = \sin y,

it should be x = n π + {(- 1)}^{n} y .

Commented by ajfour last updated on 28/May/17

w h e n w e h a v e s p e c i a l v a l u e s

o f y s u c h a s y = n (\frac{π}{2}) w e

s h o u l d p r e f e r w r i t i n g i n t h a t

m a n n e r (y o u a r e a s k i n g

j u s t i f i c a t i o n f o r) .

i f \cos x = - 1

x = 2 k π + π

o r x = 2 n π \pm π a r e e q u i v a l e n t

s i n c e f o r n = 0, y o u h a v e

x = - π, π

b o t h c a n b e o b t a i n e d w i t h k = - 1

a n d k = 0

f o r n = 1, x = π, 3 π

s a m e v a l u e s o b t a i n e d w i t h k = 0,

k = 1

w h a t i s c o m m o n t o b o t h e x p r e s s i o n s

i s t h a t x = (o d d i n t e g e r) \times π

t h i s o n l y i s n e c e s s a r y f o r

\cos x = - 1 .

A l o n g s i m i l a r l i n e s,

i f \sin x = - 1

x = 2 k π - \frac{π}{2}

o r x = n π + {(- 1)}^{n} (- \frac{π}{2}) w o u l d

a g a i n b e e q u i v a l e n t r e p r e s e n t a t i o n s .

Facebook Tweet Pin