Question

Solve for x on 0 =< x < 2pi : sin^3(2x) - sin2x= 0.

Answer 1

hi,
\[\sin ^{3}(2x) - \sin (2x) = 0\]
\[\sin ^{3}(2x) = \sin (2x)\]
\[\sin ^{2} (2x) = 1\]\[\sin (2x) = \pm \sqrt{1} =\pm1\]

when sin(2x) = +1\[\sin (2x) = \sin \frac{ \pi }{ 2 }\]
the formula for a sin angle is...\[\theta = n \pi + (-1)^{n} \alpha \ ; \ ( n \ \in \ \mathbb{Z} ) \]

which means..
\[2x = n \pi + (-1) ^{n}\frac{ \pi }{2 }\ ( n \ \in \ \mathbb{Z} )\]\[x =\frac{ n \pi }{ 2 } + (-1)^{n}\frac{ \pi }{ 4 }\]
so the values for x are...
\[x = \left\{ \frac{ \pi }{ 4 } , \frac{ 9\pi }{ 4 } , \frac{ 5 \pi }{4 } , \right\}\]

did u get it so far?

Answer 2

now for the
sin (2x) = -1

\[\sin (2x) = \sin (-\frac{ \pi }{ 2 })\]
which means ... according to the above formula..
\[2x = n \pi + (-1)^{n} (-\frac{ \pi }{ 2 })\]\[x = \frac{ n \pi }{2 } - (-1)^{n}\frac{ \pi }{4 }\]

so... the values for "x" r..
\[x = \left\{ \frac{ 3 \pi }{4 } , \frac{ 7 \pi }{4 } \right\}\]

hope this will help ya!!

Answer 3

@***[ISURU]*** Thank you!! :)