Question

Solve the following in [0,2pi)
tan2(theta)cos(theta)=cos(theta)

Answer 1

I attempted to do this problem but im not really sure if i did it right i will post what i did so far i really hope you can help me

Answer 2

sorry, i change theta be "x" it make easier :)
tan2xcosx=cosx
tan2xcosx-cosx=0
cosx(tan2x-1)=0
for zeros, cosx=0 or tan2x-1=0
* cosx=0 --->x=pi/2, 3pi/2
** tan2x-1=0
tan2x=1/2 (hmmm... not special degree for tan has value equal 1/2)

Answer 3

ok for cos would i write it theta=pi/2+2pi k and theta=3pi/2+2pi k?

Answer 4

and for tan2x-1=0 can you do it with tan2x=1 instead of zero please what would that be?

Answer 5

hello?

Answer 6

sorry, my conection very slow :(
yeaa, i mistaken just knew (above)...
tan2x-1=0
tan2x=1
tan2x=tanpi/4
tan2x=tan(k*pi+pi/4)
2x=(k*pi+pi/4
x=k*pi/2 + pi/8

tan2x=tan5pi/4
tan2x=tan(k*pi+5pi/4)
2x=k*pi+5pi/4
x=k*pi/2+5pi/8

Answer 7

thank you so would pi/2,5pi/8,3pi/2,pi/8 be the only ones in the solution range [0,2pi)

Answer 8

welcome :)

Answer 9

are those that i written above correct ?

Answer 10

i also have another question that i really need help on and if you can help me i would really appreciate it

Answer 11

for x=k*pi/2 + pi/8
if k=0, --> x=pi/8
k=1, --> x=pi/2 + pi/8 = 5pi/8
k=2, --> x=pi + pi/8 = 9pi/8
k=3, --> x=3pi/2 + pi/8 = 13pi/8
k=4,5,6,so on not satisfied in interval [0,2pi)

Answer 12

thank you can you help me with this next problem im really stuck on how to solve it

Answer 13

yeps...

Answer 14

ill write it in a new problem so then you can get another medal

Answer 15

ok