sin-3-x-cos-3-x-1-1-2-sin2x-x-0-2pi-find-x-

Question Number 63703 by kaivan.ahmadi last updated on 07/Jul/19

{s i n}^{3} x + {c o s}^{3} x = 1 - \frac{1}{2} s i n 2 x : x \in [0, 2 π] .

f i n d x

Commented by Prithwish sen last updated on 08/Jul/19

(sinx + cosx) (1 - \frac{1}{2} \sin 2 x) - (1 - \frac{1}{2} \sin 2 x) = 0

\Rightarrow \sin 2 x = 2 which is impossible

and \sin (\frac{π}{4} + x) = \sin \frac{π}{4} or \sin \frac{3 π}{4}

Commented by kaivan.ahmadi last updated on 08/Jul/19

(s i n x + c o s x - 1) (1 - \frac{1}{2} s i n 2 x) = 0 \Rightarrow

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

x = 0, \frac{π}{2}, 2 π

Answered by MJS last updated on 07/Jul/19

x = 2 \arctan t

- \frac{2 t (t - 1) (t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 1)}{{(t^{2} + 1)}^{3}} = 0

t_{1} = 0

t_{2} = 1

no other real solutions

\Rightarrow x = 0 \lor x = \frac{π}{2}

Commented by kaivan.ahmadi last updated on 08/Jul/19

1 . \frac{5 π}{2}

2 . \frac{7 π}{2}

3 . 2 π

4 . 3 π

e x c u s e m e s i r f i n d s u m o f a n s w e r s

Commented by kaivan.ahmadi last updated on 08/Jul/19

x = 2 π i s a n a n s w e r

Commented by MJS last updated on 08/Jul/19

generally x = 2 n π \lor x = \frac{π}{2} + 2 n π; n \in Z

Commented by MJS last updated on 08/Jul/19

\frac{7 π}{2} and 3 π are wrong

Commented by kaivan.ahmadi last updated on 08/Jul/19

s i r c a n y o u e x p l a i n h o w d o y o u

c h a n g e t h e v a r i a b l e ?

Commented by MJS last updated on 08/Jul/19

the substitution x = 2 \arctan t leads to

\sin x = \frac{2 t}{t^{2} + 1}

\cos x = - \frac{t^{2} - 1}{t^{2} + 1}

Commented by kaivan.ahmadi last updated on 08/Jul/19

t h a n k s s i r