Question

solve over domain -3pi/2 greater or equal to x < pi/2 for (cos2x)(cosx) - (sin2x)(sinx) = 0
trig identity: cos(alpha + beta)

Answer 1

@jim_thompson5910

Answer 2

cos 3x = 0

Answer 3

@soapia yes u are correct Cos3x = 0......
But can u write the domain again???

Answer 4

-3pi/2 greater or equal to x < pi/2

Answer 5

u want 2 say \[-3\pi /2\le x <\pi /2\]

Answer 6

yes

Answer 7

See general solution for \[\cos x = 0\] is \[x = 2n \pi \pm \pi / 2\]

So if we check the domain we get \[3x = -\pi / 2, -\pi / 2, \pi / 2\] etc.

So, \[x= -\pi / 2, -\pi / 6, \pi / 6\]

Answer 8

I think the general solution is:
\[\cos (x ) = 0\]

\[x = (2n + 1) \frac{\pi}{2} \qquad where \; \; n \in \; \mathcal{Z}\]

Answer 9

Or as @lordcyborg says:
it should be:
\[x = n \pi + \frac{\pi}{2}\]